Chapter 8 Statics and Torque

8.3 Applications of Statics and Problem-Solving Strategies

Summary

  • Discuss the applications of Statics in real life.
  • State and discuss various problem-solving strategies in Statics.

Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We begin with a discussion of problem-solving strategies specifically used for statics. Since statics is a special case of Newton’s laws, both the general problem-solving strategies and the special strategies for Newton’s laws, discussed in Chapter 4.7 Problem-Solving Strategies, still apply.

PROBLEM-SOLVING STRATEGY: STATIC EQUILIBRIUM SITUATIONS

  1. The first step is to determine whether or not the system is in static equilibrium. This condition is always the case when the acceleration of the system is zero and accelerated rotation does not occur.
  2. It is particularly important to draw a free body diagram for the system of interest. Carefully label all forces, and note their relative magnitudes, directions, and points of application whenever these are known.
  3. Solve the problem by applying either or both of the conditions for equilibrium (represented by the equations net F = 0 and  net τ =0, depending on the list of known and unknown factors. If the second condition is involved, choose the pivot point to simplify the solution. Any pivot point can be chosen, but the most useful ones cause torques by unknown forces to be zero. (Torque is zero if the force is applied at the pivot (then r = 0), or along a line through the pivot point (then θ = 0). Always choose a convenient coordinate system for projecting forces.
  4. Check the solution to see if it is reasonable by examining the magnitude, direction, and units of the answer. The importance of this last step never diminishes, although in unfamiliar applications, it is usually more difficult to judge reasonableness. These judgments become progressively easier with experience.

Now let us apply this problem-solving strategy for the pole vaulter shown in the three figures below. The pole is uniform and has a mass of 5.00 kg. In Figure 1, the pole’s cg lies halfway between the vaulter’s hands. It seems reasonable that the force exerted by each hand is equal to half the weight of the pole, or 24.5 N. This obviously satisfies the first condition for equilibrium (net F = 0). The second condition (net τ = 0) is also satisfied, as we can see by choosing the cg to be the pivot point. The weight exerts no torque about a pivot point located at the cg, since it is applied at that point and its lever arm is zero. The equal forces exerted by the hands are equidistant from the chosen pivot, and so they exert equal and opposite torques. Similar arguments hold for other systems where supporting forces are exerted symmetrically about the cg. For example, the four legs of a uniform table each support one-fourth of its weight.

In Figure 1, a pole vaulter holding a pole with its cg halfway between his hands is shown. Each hand exerts a force equal to half the weight of the pole, FR = FL = w/2. (b) The pole vaulter moves the pole to his left, and the forces that the hands exert are no longer equal. See Figure 1. If the pole is held with its cg to the left of the person, then he must push down with his right hand and up with his left. The forces he exerts are larger here because they are in opposite directions and the cg is at a long distance from either hand.

Similar observations can be made using a meter stick held at different locations along its length.

A pole vaulter is standing on the ground holding a pole with his two hands. The center of gravity of the pole is between the hands of the pole vaulter and is near the right hand of the man. The weight W is shown as an arrow downward toward center of gravity. The reactions F sub R and F sub L of the hands of the man are shown with vectors in upward direction. A free body diagram of the situation is shown on the top right side of the figure.
Figure 1. A pole vaulter holds a pole horizontally with both hands.
A pole vaulter is standing on the ground holding a pole with his two hands. The center of gravity of the pole is between the hands of the pole vaulter and is near the right hand of the man. The weight W is shown as an arrow downward toward center of gravity. The reactions F sub R and F sub L of the hands of the man are shown with vectors in upward direction. A free body diagram of the situation is shown on the top right side of the figure.
Figure 2. A pole vaulter is holding a pole horizontally with both hands. The center of gravity is near his right hand.
A pole vaulter is standing on the ground holding a pole from one side with his two hands. The centre of gravity of the pole is to the left of the pole vaulter. The weight W is shown as an arrow downward at center of gravity. The reaction F sub R is shown with a vector pointing downward from the man’s right hand and F sub L is shown with a vector in upward direction at the location of the man’s left hand. A free body diagram of the situation is shown on the top right side of the figure.
Figure 3. A pole vaulter is holding a pole horizontally with both hands. The center of gravity is to the left side of the vaulter.

If the pole vaulter holds the pole as shown in Figure 2, the situation is not as simple. The total force he exerts is still equal to the weight of the pole, but it is not evenly divided between his hands. (If FL = FR, then the torques about the cg would not be equal since the lever arms are different.) Logically, the right hand should support more weight, since it is closer to the cg. In fact, if the right hand is moved directly under the cg, it will support all the weight. This situation is exactly analogous to two people carrying a load; the one closer to the cg carries more of its weight. Finding the forces FL and FR is straightforward, as the next example shows.

If the pole vaulter holds the pole from near the end of the pole (Figure 3), the direction of the force applied by the right hand of the vaulter reverses its direction.

Example 1: What Force Is Needed to Support a Weight Held Near Its CG?

For the situation shown in Figure 2, calculate: (a) FR, the force exerted by the right hand, and (b) FL, the force exerted by the left hand. The hands are 0.900 m apart, and the cg of the pole is 0.600 m from the left hand.

Strategy

Figure 2 includes a free body diagram for the pole, the system of interest. There is not enough information to use the first condition for equilibrium (net F = 0), since two of the three forces are unknown and the hand forces cannot be assumed to be equal in this case. There is enough information to use the second condition for equilibrium (net τ = 0) if the pivot point is chosen to be at either hand, thereby making the torque from that hand zero. We choose to locate the pivot at the left hand in this part of the problem, to eliminate the torque from the left hand.

Solution for (a)

There are now only two nonzero torques, those from the gravitational force (τw) and from the push or pull of the right hand (τR). Stating the second condition in terms of clockwise and counterclockwise torques,

[latex]\text{net}\tau_{\text{cw}}=-\text{net}~\tau_{\text{ccw}}.[/latex]

or the algebraic sum of the torques is zero.

Here this is

[latex]\tau_{\text{R}}=-\tau_{\text{w}}[/latex]

since the weight of the pole creates a counterclockwise torque and the right hand counters with a clockwise torque. Using the definition of torque, τ = rF sin θ, noting that θ = 90°, and substituting known values, we obtain

[latex](0.900~\text{m})(F_{\text{R}})=(0.600~\text{m})(mg)[/latex]

Thus,

[latex]\begin{align*} F_{\text{R}} &= (0.667)(5.00~\text{kg})(9.80~\text{m/s}^2) \\ &= 32.7~\text{N.} \end{align*}[/latex]

Solution for (b)

The first condition for equilibrium is based on the free body diagram in the figure. This implies that by Newton’s second law:

[latex]F_{\text{L}}+F_{\text{R}}-mg=0[/latex]

From this we can conclude:

[latex]F_{\text{L}}+F_{\text{R}}=w=mg[/latex]

Solving for FL, we obtain

[latex]\begin{align*} F_{\text{L}} &= mg-F_{\text{R}} \\ &= mg-32.7~\text{N} \\ &= (5.00~\text{kg})(9.80~\text{m/s}^2)-32.7~\text{N} \\&= 16.3~\text{N.} \end{align*}[/latex]

Discussion

FLis seen to be exactly half of FR, as we might have guessed, since FL is applied twice as far from the cg as FR.

If the pole vaulter holds the pole as he might at the start of a run, shown in Figure 3, the forces change again. Both are considerably greater, and one force reverses direction.

TAKE-HOME EXPERIMENT

This is an experiment to perform while standing in a bus or a train. Stand facing sideways. How do you move your body to readjust the distribution of your mass as the bus accelerates and decelerates? Now stand facing forward. How do you move your body to readjust the distribution of your mass as the bus accelerates and decelerates? Why is it easier and safer to stand facing sideways rather than forward? Note: For your safety (and those around you), make sure you are holding onto something while you carry out this activity!

PHET EXPLORATIONS: BALANCING ACT

Play with objects on a teeter totter to learn about balance. Test what you’ve learned by trying the Balance Challenge game.

image
Figure 4. Balancing Act

Summary

  • Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We have discussed the problem-solving strategies specifically useful for statics. Statics is a special case of Newton’s laws, both the general problem-solving strategies and the special strategies for Newton’s laws, discussed in Chapter 4.7 Problem-Solving Strategies, still apply.

Glossary

static equilibrium
equilibrium in which the acceleration of the system is zero and accelerated rotation does not occur

Conceptual Questions

1: When visiting some countries, you may see a person balancing a load on the head. Explain why the center of mass of the load needs to be directly above the person’s neck vertebrae.

Problems & Exercises

1: Suppose a horse leans against a wall as in Figure 10. Calculate the force exerted on the wall assuming that force is horizontal while using the data in the schematic representation of the situation. Note that the force exerted on the wall is equal in magnitude and opposite in direction to the force exerted on the horse, keeping it in equilibrium. The total mass of the horse and rider is 500 kg. Take the data to be accurate to three digits.

In part a, a horse is standing next to a wall with its legs crossed. A sleepy-looking rider is leaning against the wall. Part b is a drawing of the same horse from a rear view, but this time with no rider. The horse is crossing its rear legs, and its rump is leaning against the wall. The reaction of the wall F is acting on the horse at a height one point two meters above the ground. The weight of the horse is acting at its center of gravity near the base of the tail. The center of gravity is one point four meters above the ground. The line of action of weight is zero point three five meters away from the feet of the horse.
Figure 10.

2: Two children of mass 20.0 kg and 30.0 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the children are separated by a distance of 3.00 m, at what distance from the pivot point is the small child sitting in order to maintain the balance?

3: (a) Calculate the magnitude and direction of the force on each foot of the horse in Figure 10 (two are on the ground), assuming the center of mass of the horse is midway between the feet. The total mass of the horse and rider is 500kg. (b) What is the minimum coefficient of friction between the hooves and ground? Note that the force exerted by the wall is horizontal.

4: A person carries a plank of wood 2.00 m long with one hand pushing down on it at one end with a force F1 and the other hand holding it up at .500 m from the end of the plank with force F2. If the plank has a mass of 20.0 kg and its center of gravity is at the middle of the plank, what are the magnitudes of the forces F1 and F2?

5: A 17.0-m-high and 11.0-m-long wall under construction and its bracing are shown in Figure 11. The wall is in stable equilibrium without the bracing but can pivot at its base. Calculate the force exerted by each of the 10 braces if a strong wind exerts a horizontal force of 650 N on each square meter of the wall. Assume that the net force from the wind acts at a height halfway up the wall and that all braces exert equal forces parallel to their lengths. Neglect the thickness of the wall.

A seventeen meter high wall is standing on the ground with ten braces to support it. At the base of the figure a brown colored ground is visible. Only one brace is visible from a side. A brace makes an angle of thirty five degree with the wall. The point of contact of the brace is eight point five meters high. You have to find the force exerted by this brace on the wall to support.
Figure 11.

6: (a) What force must be exerted by the wind to support a 2.50-kg chicken in the position shown in Figure 12? (b) What is the ratio of this force to the chicken’s weight? (c) Does this support the contention that the chicken has a relatively stable construction?

A chicken is trying to balance on its left foot, which is 9 point zero centimeters to the right of the chicken. The force of the wind is blowing from the left toward the chicken’s center of gravity c g, which is 20 cm above the ground. The weight of the chicken w is acting at the center of gravity.
Figure 12.

7: Suppose the weight of the drawbridge in Figure 13 is supported entirely by its hinges and the opposite shore, so that its cables are slack. (a) What fraction of the weight is supported by the opposite shore if the point of support is directly beneath the cable attachments? (b) What is the direction and magnitude of the force the hinges exert on the bridge under these circumstances? The mass of the bridge is 2500 kg.

A small drawbridge is shown. There is one vertical and one horizontal wooden plank. The left end of the horizontal plank is attached to the vertical plank near its middle. At the point of contact, a hinge is shown. A wire is tied to the right end of the horizontal end, is passed over the top of the vertical plank and is connected to a pulley. The angle made by the wire with the horizontal plank is forty degrees. The reaction F at the hinge is inclined at an angle theta.
Figure 13. A small drawbridge, showing the forces on the hinges (F), its weight (w), and the tension in its wires (T).

8: Suppose a 900-kg car is on the bridge in Figure 13 with its center of mass halfway between the hinges and the cable attachments. (The bridge is supported by the cables and hinges only.) (a) Find the force in the cables. (b) Find the direction and magnitude of the force exerted by the hinges on the bridge.

9: A sandwich board advertising sign is constructed as shown in Figure 14. The sign’s mass is 8.00 kg. (a) Calculate the tension in the chain assuming no friction between the legs and the sidewalk. (b) What force is exerted by each side on the hinge?

A sandwich board advertising sign is in form of a triangle. The base of the triangle is one point one zero meters. The other two sides are connected with a hinge at the top. A horizontal chain is connected to the two legs at zero point five zero meters below the hinge. The height of the hinge above the base is one point three zero meters. The centers of the gravity of the two legs are shown at their midpoints. The figure is labeled at uniform board with c g at the center.
Figure 14. A sandwich board advertising sign demonstrates tension.

10: (a) What minimum coefficient of friction is needed between the legs and the ground to keep the sign in Figure 14 in the position shown if the chain breaks? (b) What force is exerted by each side on the hinge?

11: A gymnast is attempting to perform splits. From the information given in Figure 15, calculate the magnitude and direction of the force exerted on each foot by the floor.

A gymnast with two pompoms in her hands is shown. One of the hand is horizontal toward left and the other is vertical. The gymnast is attempting to perform a full split. The span of her legs is one point eight meters, and the distance of one foot from the center of gravity is zero point nine meters. The weight of the girl is labeled as seven hundred newtons. The vertical distance of one foot from the center of gravity is zero point three zero meter.
Figure 15. A gymnast performs full split. The center of gravity and the various distances from it are shown.

 

12: To get up on the roof, a person (mass 70.0 kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes of the forces on the ladder at the top and bottom?

13: In Figure 3, the cg of the pole held by the pole vaulter is 2.00 m from the left hand, and the hands are 0.700 m apart. Calculate the force exerted by (a) his right hand and (b) his left hand. (c) If each hand supports half the weight of the pole in Figure 1, show that the second condition for equilibrium (net τ = 0) is satisfied for a pivot other than the one located at the center of gravity of the pole. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium described above.

 

Problems & Exercises

1: To get up on the roof, a person (mass 70.0 kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes of the forces on the ladder at the top and bottom?

2: In Figure 3, the cg of the pole held by the pole vaulter is 2.00 m from the left hand, and the hands are 0.700 m apart. Calculate the force exerted by (a) his right hand and (b) his left hand. (c) If each hand supports half the weight of the pole in Figure 1, show that the second condition for equilibrium (net τ = 0) is satisfied for a pivot other than the one located at the center of gravity of the pole. Explicitly show how you follow the steps in the Problem-Solving Strategy for static equilibrium described above.

Solutions

Problems & Exercises

1: Fwall = 1.43 x 103 N

3: (a) 2.55 x 103 N, 16.3o to the left of vertical (i.e, toward the wall) (b) 0.292

5: FB = 2.12 x 104 N

7: (a) 0.167, or about one-sixth of the weight is supported by the opposite shore. (b) F = 2.0 x 104 N straight up.

9: (a) 21.6 N (b) 21.6 N

11: 350 N directly upwards

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