what is the ratio of the minimum tension in cord 2 to the weight of the package?

Learning Objectives

Past the end of this section, you will be able to:

• Ascertain normal and tension forces.
• Apply Newton'south laws of movement to solve issues involving a multifariousness of forces.
• Use trigonometric identities to resolve weight into components.

Forces are given many names, such as push, pull, thrust, lift, weight, friction, and tension. Traditionally, forces have been grouped into several categories and given names relating to their source, how they are transmitted, or their effects. The most important of these categories are discussed in this department, together with some interesting applications. Farther examples of forces are discussed later in this text.

Normal Forcefulness

Weight (as well chosen forcefulness of gravity) is a pervasive force that acts at all times and must be counteracted to proceed an object from falling. You definitely notice that you lot must back up the weight of a heavy object past pushing upward on it when you hold it stationary, every bit illustrated in Effigy 1(a). But how practise inanimate objects like a tabular array support the weight of a mass placed on them, such as shown in Figure 1(b)? When the bag of dog food is placed on the tabular array, the table actually sags slightly under the load. This would be noticeable if the load were placed on a card table, but fifty-fifty rigid objects deform when a force is applied to them. Unless the object is deformed beyond its limit, it will exert a restoring strength much similar a deformed spring (or trampoline or diving lath). The greater the deformation, the greater the restoring strength. Then when the load is placed on the tabular array, the tabular array sags until the restoring strength becomes equally large equally the weight of the load. At this point the net external force on the load is zero. That is the state of affairs when the load is stationary on the table. The table sags speedily, and the sag is slight so we do not notice information technology. Only it is similar to the sagging of a trampoline when you lot climb onto it.

Figure 1. (a) The person belongings the bag of canis familiaris food must supply an upward force F_hand equal in magnitude and opposite in management to the weight of the food w. (b) The card table sags when the canis familiaris nutrient is placed on it, much like a potent trampoline. Elastic restoring forces in the tabular array grow as it sags until they supply a force N equal in magnitude and opposite in direction to the weight of the load.

We must conclude that whatever supports a load, be it animate or not, must supply an upward force equal to the weight of the load, as we assumed in a few of the previous examples. If the force supporting a load is perpendicular to the surface of contact between the load and its support, this force is defined to be a normal force and here is given the symbol North. (This is non the unit for force N.) The word normal means perpendicular to a surface. The normal force can be less than the object's weight if the object is on an incline, as y'all will see in the next example.

Common Misconception: Normal Force (Due north) vs. Newton (Due north)

In this department we accept introduced the quantity normal strength, which is represented by the variable N. This should not be dislocated with the symbol for the newton, which is besides represented by the letter N. These symbols are particularly of import to distinguish considering the units of a normal strength (N) happen to be newtons (N). For example, the normal force Northward that the floor exerts on a chair might be North = 100 N. I important difference is that normal force is a vector, while the newton is simply a unit of measurement. Be careful not to confuse these letters in your calculations! You volition encounter more similarities amidst variables and units every bit you proceed in physics. Another example of this is the quantity piece of work (W) and the unit watts (W).

Case ane. Weight on an Incline, a Two-Dimensional Problem

Consider the skier on a slope shown in Figure 2. Her mass including equipment is 60.0 kg. (a) What is her acceleration if friction is negligible? (b) What is her acceleration if friction is known to be 45.0 N?

Figure two. Since movement and friction are parallel to the slope, it is most user-friendly to project all forces onto a coordinate organization where i axis is parallel to the slope and the other is perpendicular (axes shown to left of skier). N is perpendicular to the slope and f is parallel to the slope, but w has components along both axes, namely w_⊥ and [latex]\textbf{westward}_{\parallel}[/latex]. N is equal in magnitude to due west_⊥ , then that at that place is no motion perpendicular to the slope, simply f is less than westward∥, so that at that place is a downslope acceleration (along the parallel axis).

Strategy

This is a two-dimensional trouble, since the forces on the skier (the system of interest) are not parallel. The approach we take used in two-dimensional kinematics also works very well here. Choose a user-friendly coordinate arrangement and project the vectors onto its axes, creating two connected one-dimensional problems to solve. The most user-friendly coordinate organisation for motion on an incline is one that has i coordinate parallel to the slope and one perpendicular to the gradient. (Think that motions forth mutually perpendicular axes are independent.) We utilize the symbols ⊥ and ∥ to represent perpendicular and parallel, respectively. This selection of axes simplifies this type of problem, because there is no motion perpendicular to the gradient and because friction is always parallel to the surface between ii objects. The simply external forces acting on the organisation are the skier's weight, friction, and the back up of the slope, respectively labeled w, f, and North in Figure 2. N is always perpendicular to the slope, and f is parallel to information technology. Only w is not in the direction of either centrality, and so the first step we take is to project it into components forth the chosen axes, defining due west _∥ to be the component of weight parallel to the gradient and w _⊥ the component of weight perpendicular to the slope. Once this is washed, we tin consider the ii separate problems of forces parallel to the slope and forces perpendicular to the slope.

Solution

The magnitude of the component of the weight parallel to the slope is [latex]{westward}_{\parallel }={w} \sin({ 25}^{\circ}) = mg\sin ({ 25}^{\circ})[/latex], and the magnitude of the component of the weight perpendicular to the slope is [latex]{w}_{\perp}={due west}\cos({25}^{\circ}) = mg\cos({25}^{\circ})[/latex].

(a) Neglecting friction. Since the acceleration is parallel to the gradient, we need merely consider forces parallel to the slope. (Forces perpendicular to the slope add to zero, since there is no acceleration in that direction.) The forces parallel to the slope are the amount of the skier's weight parallel to the slope due west _∥ and friction f. Using Newton'due south second constabulary, with subscripts to announce quantities parallel to the slope,

[latex]{a}_{\parallel }=\frac{{F}_{\text{net}\parallel }}{m}[/latex]

where [latex]{F}_{\text{cyberspace}\parallel}={west}_{\parallel}=mg\sin({25^{\circ}})[/latex], assuming no friction for this part, so that

[latex]a_{\parallel}=\frac{{F}_{\text{net}\parallel}}{thousand}=\frac{{mg}\sin({25}^{\circ})}{g}=g\sin({25}^{\circ})[/latex]

(9.98 m/southward²)(0.4226) = four.14 m/s²

is the acceleration.

(b) Including friction. We now have a given value for friction, and we know its direction is parallel to the slope and information technology opposes motion between surfaces in contact. And so the cyberspace external forcefulness is now

[latex]{F}_{\text{cyberspace}\parallel }={w}_{\parallel }-f[/latex]

and substituting this into Newton'south 2d law, [latex]{a}_{\parallel}=\frac{{F}_{\text{net}\parallel }}{m}[/latex], gives

[latex]{a}_{\parallel}=\frac{{F}_{\text{net}\parallel}}{m}=\frac{{w}_{\parallel}-f}{m}=\frac{{mg}\sin({25}^{\circ})-f}{m}[/latex].

We substitute known values to obtain

[latex]{a}_{\parallel }=\frac{(threescore.0\text{ kg})(9.80\text{ m/s}^{two})(0.4226)-45.0\text{ Northward}}{sixty.0\text{ kg}}[/latex]

which yields

[latex]a_{\parallel}= 3.39\text{ one thousand/south}^{2}[/latex]

which is the acceleration parallel to the incline when in that location is 45.0 N of opposing friction.

Word

Since friction always opposes motion between surfaces, the dispatch is smaller when there is friction than when there is none. In fact, it is a full general result that if friction on an incline is negligible, then the acceleration down the incline isa =g sinθ, regardless of mass. This is related to the previously discussed fact that all objects fall with the same acceleration in the absence of air resistance. Similarly, all objects, regardless of mass, slide downwardly a frictionless incline with the same acceleration (if the bending is the aforementioned).

Resolving Weight into Components

Figure 3. An object rests on an incline that makes an angle θ with the horizontal.

When an object rests on an incline that makes an angle θ with the horizontal, the strength of gravity acting on the object is divided into two components: a force interim perpendicular to the plane, westward_⊥ , and a force interim parallel to the plane,[latex]\textbf{w}_{\parallel}[/latex]. The perpendicular strength of weight, w_⊥, is typically equal in magnitude and contrary in management to the normal force, N. The forcefulness interim parallel to the plane, [latex]\textbf{w}_{\parallel}[/latex], causes the object to accelerate down the incline. The force of friction, f, opposes the motion of the object, so information technology acts upwardly forth the plane. It is of import to exist careful when resolving the weight of the object into components. If the angle of the incline is at an angle θ to the horizontal, then the magnitudes of the weight components are

[latex]w_{\parallel}=westward \sin{\theta} = mg \sin{\theta}[/latex]

and

[latex]w_{\perp}=west \cos{\theta} = mg \cos{\theta}[/latex]

Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, draw the right triangle formed past the three weight vectors. Notice that the angle θ of the incline is the same as the angle formed between w and w_⊥. Knowing this property, you can use trigonometry to determine the magnitude of the weight components:

[latex]\cos({\theta})=\frac{{w}_{\perp}}{w}[/latex]

[latex]w_{\perp} = w \cos{\theta} = mg \cos({\theta})[/latex]

[latex]\sin{(\theta)}=\frac{{westward}_{\parallel}}{w}[/latex]

[latex]w_{\parallel} = w \sin{\theta} = mg \sin({\theta})[/latex]

Take-Home Experiment: Force Parallel

To investigate how a force parallel to an inclined aeroplane changes, find a rubber band, some objects to hang from the end of the rubber band, and a board you tin position at different angles. How much does the rubber ring stretch when you hang the object from the finish of the board? At present place the board at an angle so that the object slides off when placed on the board. How much does the rubber band extend if it is lined upwardly parallel to the lath and used to hold the object stationary on the board? Try two more angles. What does this testify?

Tension

A tension is a force along the length of a medium, especially a force carried past a flexible medium, such as a rope or cablevision. The word "tension" comes from a Latin give-and-take pregnant "to stretch." Not coincidentally, the flexible cords that bear musculus forces to other parts of the body are called tendons. Whatever flexible connector, such as a cord, rope, chain, wire, or cable, can exert pulls only parallel to its length; thus, a force carried by a flexible connector is a tension with direction parallel to the connector. Information technology is important to sympathize that tension is a pull in a connector. In contrast, consider the phrase: "You can't push a rope." The tension strength pulls outward forth the two ends of a rope. Consider a person holding a mass on a rope as shown in Figure 4.

Figure 4. When a perfectly flexible connector (i requiring no strength to curve it) such as this rope transmits a force T, that force must be parallel to the length of the rope, every bit shown. The pull such a flexible connector exerts is a tension. Note that the rope pulls with equal forcefulness just in opposite directions on the hand and the supported mass (neglecting the weight of the rope). This is an case of Newton's tertiary law. The rope is the medium that carries the equal and reverse forces between the two objects. The tension anywhere in the rope betwixt the mitt and the mass is equal. In one case you take determined the tension in 1 location, you have determined the tension at all locations forth the rope.

Tension in the rope must equal the weight of the supported mass, equally nosotros tin prove using Newton'southward second police. If the 5.00-kg mass in the effigy is stationary, then its dispatch is zero, and thus F_net= 0. The only external forces acting on the mass are its weight w and the tension T supplied past the rope. Thus,

F _net=T−west= 0,

where T and w are the magnitudes of the tension and weight and their signs signal direction, with upward existence positive here. Thus, just as you would expect, the tension equals the weight of the supported mass:

T = west = mg.

For a five.00-kg mass, then (neglecting the mass of the rope) we come across that

T = mg = (5.00 kg)(nine.80 1000/s²) = 49.0 N

If we cut the rope and insert a spring, the spring would extend a length corresponding to a force of 49.0 Due north, providing a direct ascertainment and measure out of the tension force in the rope. Flexible connectors are often used to transmit forces around corners, such every bit in a infirmary traction system, a finger articulation, or a bicycle brake cable. If there is no friction, the tension is transmitted undiminished. But its direction changes, and it is e'er parallel to the flexible connector. This is illustrated in Figure 5 (a) and (b).

Effigy v. (a) Tendons in the finger carry forcefulness T from the muscles to other parts of the finger, usually irresolute the force's management, but not its magnitude (the tendons are relatively friction complimentary). (b) The brake cablevision on a bicycle carries the tension T from the handlebars to the brake mechanism. Again, the direction but not the magnitude of T is changed.

Example 2. What Is the Tension in a Tightrope?

Calculate the tension in the wire supporting the 70.0-kg tightrope walker shown in Figure half dozen.

Effigy 6. The weight of a tightrope walker causes a wire to sag by 5.0 degrees. The system of involvement hither is the signal in the wire at which the tightrope walker is standing.

Strategy

As yous tin come across in the figure, the wire is non perfectly horizontal (it cannot exist!), only is bent under the person's weight. Thus, the tension on either side of the person has an upwardly component that tin can support his weight. As usual, forces are vectors represented pictorially by arrows having the same directions as the forces and lengths proportional to their magnitudes. The system is the tightrope walker, and the just external forces acting on him are his weight west and the 2 tensions T_Fifty (left tension) and T_R (right tension), every bit illustrated. It is reasonable to neglect the weight of the wire itself. The net external force is zero since the organisation is stationary. A little trigonometry tin now be used to find the tensions. Ane determination is possible at the outset—we tin can see from part (b) of the figure that the magnitudes of the tensions T _L and T _R must be equal. This is considering at that place is no horizontal dispatch in the rope, and the simply forces acting to the left and right are T _L and T _R . Thus, the magnitude of those forces must exist equal so that they cancel each other out.

Whenever nosotros accept two-dimensional vector issues in which no 2 vectors are parallel, the easiest method of solution is to pick a convenient coordinate system and projection the vectors onto its axes. In this case the best coordinate system has one centrality horizontal and the other vertical. We phone call the horizontal the 10-axis and the vertical the y-axis.

Solution

First, we need to resolve the tension vectors into their horizontal and vertical components. Information technology helps to describe a new free-body diagram showing all of the horizontal and vertical components of each force acting on the system.

Figure 7. When the vectors are projected onto vertical and horizontal axes, their components along those axes must add to zero, since the tightrope walker is stationary. The small angle results in T being much greater than w.

Consider the horizontal components of the forces (denoted with a subscript x):

F _netx = T _Lx − T _Rx .

The net external horizontal forcefulness F _{internet x} = 0, since the person is stationary. Thus,

F _cyberspacex = T _Fiftyten − T _Rx

T _Lx =T _Rx

Now, observe Figure seven. You can utilize trigonometry to decide the magnitude of T _Fifty and T _R. Find that:

[latex]\begin{assortment}{lll}{\cos}\left(v.0^{\circ}\correct)& =& \frac{{T}_{\text{L}10}}{{T}_{\text{L}}}\\ {T}_{\text{L}x}& =& {T}_{\text{50}}\cos\left(5.0^{\circ}\correct)\\ \cos\left(five.0^{\circ}\right)& =& \frac{{T}_{\text{R}x}}{{T}_{\text{R}}}\\ {T}_{\text{R}x}& =& {T}_{\text{R}}\cos\left(5.0^{\circ}\right)\end{array}[/latex]

Equating T _{L x} and T _Rx :

[latex]{T}_{\text{L}}\cos({5.0}^{\circ})={T}_{\text{R}}\cos({5.0}^{\circ})[/latex].

Thus,

T _L=T _R=T,

equally predicted. Now, because the vertical components (denoted by a subscript y), we can solve for T. Over again, since the person is stationary, Newton's second police implies that net F _y =0. Thus, equally illustrated in the free-body diagram in Figure 7,

F _{net y} =T _Ly +T _Ry −w= 0.

Observing Figure 7, we can utilise trigonometry to determine the relationship between T _Ly , T _Ry , and T. Every bit we adamant from the analysis in the horizontal direction, T _L=T _R=T:

[latex]\brainstorm{array}{lll}\sin\left(5.0^{\circ}\right)& =& \frac{{T}_{\text{L}y}}{{T}_{\text{L}}}\\ {T}_{\text{Fifty}y}={T}_{\text{L}}\sin\left(v.0^{\circ} \correct)& =& T\sin\left(five.0^{\circ}\correct)\\ \sin\left(5.0^{\circ}\right)& =& \frac{{T}_{\text{R}y}}{{T}_{\text{R}}}\\ {T}_{\text{R}y}={T}_{\text{R}}\sin\left(5.0^{\circ}\right)& =& T\sin\left(5.0^{\circ}\right)\cease{array}[/latex].

Now, we can substitute the values for T _50y and T _Ry , into the net force equation in the vertical direction:

[latex]\brainstorm{array}{lll}{F}_{\text{net}y}& =& {T}_{\text{L}y}+{T}_{\text{R}y}-w=0\\ {F}_{\text{internet}y}& =& T\sin\left(5.0^{\circ}\right)+T\sin\left(v.0^{\circ}\correct)-w=0\\ 2T\sin\left(v.0^{\circ}\right)-due west& =& 0\\ 2T\sin\left(v.0^{\circ} \right)& =& w\stop{assortment}[/latex]

and

[latex]T=\frac{w}{2\sin({5.0}^{\circ})}=\frac{\text{mg}}{two\sin({5.0}^{\circ})}[/latex],

so that

[latex]T=\frac{(70.0\text{ kg})(9.80\text{ k/s}^{2})}{two(0.0872)}[/latex],

and the tension is

T= 3900 N.

Discussion

Note that the vertical tension in the wire acts equally a normal force that supports the weight of the tightrope walker. The tension is almost six times the 686-N weight of the tightrope walker. Since the wire is nearly horizontal, the vertical component of its tension is merely a small fraction of the tension in the wire. The large horizontal components are in reverse directions and abolish, and and then near of the tension in the wire is not used to support the weight of the tightrope walker.

If we wish to create a very large tension, all we have to practise is exert a force perpendicular to a flexible connector, equally illustrated in Figure 8. Equally we saw in the last example, the weight of the tightrope walker acted every bit a force perpendicular to the rope. We saw that the tension in the roped related to the weight of the tightrope walker in the following way:

[latex]T=\frac{w}{2\sin({\theta})}[/latex].

Nosotros tin can extend this expression to describe the tension T created when a perpendicular forcefulness (F_⊥) is exerted at the middle of a flexible connector:

[latex]T=\frac{{F}_{\perp }}{two\sin({\theta})}[/latex].

Note that θ is the angle between the horizontal and the aptitude connector. In this case, T becomes very large as θ approaches zero. Even the relatively small weight of any flexible connector will cause it to sag, since an space tension would event if it were horizontal (i.e., θ= 0 and sinθ = 0). (Run into Effigy eight.)

Effigy 8. Nosotros can create a very big tension in the chain by pushing on it perpendicular to its length, as shown. Suppose nosotros wish to pull a car out of the mud when no tow truck is available. Each time the machine moves forward, the concatenation is tightened to proceed it every bit about directly as possible. The tension in the chain is given by [latex]T=\frac{{F}_{\perp }}{ii\sin({\theta})}[/latex]; since θ is modest, T is very big. This state of affairs is analogous to the tightrope walker shown in Figure half dozen, except that the tensions shown here are those transmitted to the auto and the tree rather than those acting at the point where F_⊥ is applied.

Figure 9. Unless an space tension is exerted, any flexible connector—such as the chain at the bottom of the film—will sag under its own weight, giving a characteristic curve when the weight is evenly distributed along the length. Suspension bridges—such as the Gilt Gate Bridge shown in this image—are substantially very heavy flexible connectors. The weight of the bridge is evenly distributed along the length of flexible connectors, ordinarily cables, which have on the characteristic shape. (credit: Leaflet, Wikimedia Eatables)

Extended Topic: Real Forces and Inertial Frames

There is some other distinction amid forces in addition to the types already mentioned. Some forces are real, whereas others are not. Real forces are those that have some physical origin, such equally the gravitational pull. Contrastingly, fictitious forces are those that arise simply considering an observer is in an accelerating frame of reference, such as one that rotates (like a merry-go-round) or undergoes linear dispatch (like a car slowing down). For case, if a satellite is heading due north to a higher place Earth's northern hemisphere, then to an observer on Earth it will appear to experience a force to the west that has no physical origin. Of grade, what is happening here is that Earth is rotating toward the eastward and moves due east under the satellite. In Earth's frame this looks like a westward strength on the satellite, or information technology can be interpreted as a violation of Newton'due south first law (the law of inertia). An inertial frame of reference is one in which all forces are real and, equivalently, one in which Newton'south laws have the simple forms given in this chapter.

Earth'south rotation is slow enough that Earth is nearly an inertial frame. Yous ordinarily must perform precise experiments to observe fictitious forces and the slight departures from Newton's laws, such every bit the outcome but described. On the big calibration, such every bit for the rotation of conditions systems and bounding main currents, the effects can be easily observed.

The crucial gene in determining whether a frame of reference is inertial is whether it accelerates or rotates relative to a known inertial frame. Unless stated otherwise, all phenomena discussed in this text are considered in inertial frames.

All the forces discussed in this section are real forces, but there are a number of other existent forces, such every bit lift and thrust, that are not discussed in this section. They are more than specialized, and information technology is not necessary to talk over every type of forcefulness. It is natural, however, to inquire where the basic simplicity nosotros seek to discover in physics is in the long list of forces. Are some more basic than others? Are some dissimilar manifestations of the same underlying force? The answer to both questions is yep, as will exist seen in the next (extended) department and in the treatment of mod physics later in the text.

PhET Explorations: Forces in 1 Dimension

Explore the forces at work when you try to push a filing cabinet. Create an practical force and see the resulting friction forcefulness and full force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. fourth dimension. View a free-torso diagram of all the forces (including gravitational and normal forces).

Click to download. Run using Java.

Section Summary

• When objects rest on a surface, the surface applies a force to the object that supports the weight of the object. This supporting strength acts perpendicular to and abroad from the surface. It is called a normal strength, Northward.
• When objects rest on a non-accelerating horizontal surface, the magnitude of the normal force is equal to the weight of the object:

  N = mg

• When objects residuum on an inclined plane that makes an angle θ with the horizontal surface, the weight of the object can be resolved into components that act perpendicular ([latex]{\mathbf{\text{w}}}_{\perp}[/latex]) and parallel ([latex]{\mathbf{\text{due west}}}_{\parallel}[/latex]) to the surface of the plane. These components can be calculated using:
  • [latex]{w}_{\parallel}=due west\sin({\theta})=mg\sin({\theta})[/latex]

  • [latex]{w}_{\perp}=west\cos({\theta})=mg\cos({\theta})[/latex]

• The pulling force that acts forth a stretched flexible connector, such as a rope or cable, is called tension, T. When a rope supports the weight of an object that is at balance, the tension in the rope is equal to the weight of the object:

  T = mg.

• In any inertial frame of reference (ane that is not accelerated or rotated), Newton's laws have the simple forms given in this chapter and all forces are real forces having a physical origin.

Conceptual Questions

1. If a leg is suspended by a traction setup as shown in Figure 9, what is the tension in the rope?

Figure nine. A leg is suspended by a traction system in which wires are used to transmit forces. Frictionless pulleys change the direction of the forcefulness T without irresolute its magnitude.

2. In a traction setup for a cleaved os, with pulleys and rope available, how might nosotros be able to increment the force along the tibia using the aforementioned weight? (See Figure ix.) (Notation that the tibia is the shin os shown in this image.)

Bug & Exercises

1. Two teams of nine members each appoint in a tug of war. Each of the offset team'due south members has an average mass of 68 kg and exerts an average strength of 1350 Due north horizontally. Each of the 2d team's members has an average mass of 73 kg and exerts an average strength of 1365 N horizontally. (a) What is magnitude of the dispatch of the 2 teams? (b) What is the tension in the section of rope betwixt the teams?

2. What force does a trampoline have to employ to a 45.0-kg gymnast to accelerate her direct up at vii.50 chiliad/due south²? Notation that the answer is contained of the velocity of the gymnast—she can be moving either up or down, or exist stationary.

3. (a) Summate the tension in a vertical strand of spider web if a spider of mass 8.00 × 10^-5 hangs motionless on information technology. (b) Calculate the tension in a horizontal strand of spider web if the same spider sits motionless in the heart of it much like the tightrope walker in Figure half dozen. The strand sags at an angle of 12º below the horizontal. Compare this with the tension in the vertical strand (discover their ratio).

4. Suppose a 60.0-kg gymnast climbs a rope. (a) What is the tension in the rope if he climbs at a constant speed? (b) What is the tension in the rope if he accelerates upward at a rate of 1.50 thousand/due south²?

5. Show that, as stated in the text, a force [latex]{\mathbf{\text{F}}}_{\perp}[/latex] exerted on a flexible medium at its center and perpendicular to its length (such equally on the tightrope wire in Figure half-dozen) gives rise to a tension of magnitude [latex]T=\frac{{F}_{\perp }}{2\sin(\theta)}[/latex].

6. Consider the baby being weighed in Figure 10. (a) What is the mass of the kid and handbasket if a scale reading of 55 N is observed? (b) What is the tension T ₁ in the cord attaching the infant to the scale? (c) What is the tension T ₂ in the string attaching the scale to the ceiling, if the scale has a mass of 0.500 kg? (d) Depict a sketch of the state of affairs indicating the system of interest used to solve each part. The masses of the cords are negligible.

Figure ten. A baby is weighed using a leap scale.

Glossary

inertial frame of reference:

a coordinate system that is not accelerating; all forces acting in an inertial frame of reference are real forces, every bit opposed to fictitious forces that are observed due to an accelerating frame of reference

normal forcefulness:

the strength that a surface applies to an object to support the weight of the object; acts perpendicular to the surface on which the object rests

tension:

the pulling force that acts forth a medium, peculiarly a stretched flexible connector, such as a rope or cable; when a rope supports the weight of an object, the strength on the object due to the rope is called a tension force

Selected Solutions to Problems & Exercises

i. (a) 0.eleven m/s^two (b) ane.2 × 10⁴ N

3. (a) seven.84 × 10^-4 N (b) 1.89 × 10^-three North. This is 2.41 times the tension in the vertical strand.

v.  Newton's second law practical in vertical direction gives

[latex]{F}_{y}=F - ii T\sin{\theta}=0[/latex]

[latex]{F}=ii T\sin{\theta}[/latex]

[latex]T=\frac{F}{2\sin{\theta}}[/latex].

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Source: https://courses.lumenlearning.com/physics/chapter/4-5-normal-tension-and-other-examples-of-forces/

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