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Chapter 9: Q9-13Q (page 230)

Why is it more difficult to do sit-ups when your knees are bent than when your legs are stretched out?

Short Answer

Expert verified

It is difficult to do sit-ups when your knees are bent because less opposing torque is produced by the lower half of the body.

Step by step solution

01

Understanding torque exerted on the body

In this problem, one of the essential points is that during the sit-up, the abdomen muscle will exert a torque to rotate the body upward, away from the ground.

02

Explaining the difficulty in doing sit-ups when your knees are bent than when your legs are stretched out

While doing a sit-up, the upper half of the body feels the force of gravity which pulls the body down toward the ground.

So, the lower half of the body applies some torque that counters the torque produced by the force of gravity on the upper half body, making sit-ups slightly easier.

If the legs are bent during sit-ups, the lever arm for the lower part of the body is short. So, the opposing torque will reduce, and this will make sit-ups difficult.

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Most popular questions from this chapter

Place yourself facing the edge of an open door. Position your feet astride the door with your nose and abdomen touching the door’s edge. Try to rise on your tiptoes. Why can’t this be done?


In a mountain-climbing technique called the “Tyrolean traverse,” a rope is anchored on both ends (to rocks or strong trees) across a deep chasm, and then a climber traverses the rope while attached by a sling as in Fig. 9–91. This technique generates tremendous forces in the rope and anchors, so a basic understanding of physics is crucial for safety. A typical climbing rope can undergo a tension force of perhaps 29 kN before breaking, and a “safety factor” of 10 is usually recommended. The length of rope used in the Tyrolean traverse must allow for some “sag” to remain in the recommended safety range. Consider a 75-kg climber at the center of a Tyrolean traverse, spanning a 25-m chasm. (a) To be within its recommended safety range, what minimum distance x must the rope sag? (b) If the Tyrolean traverse is set up incorrectly so that the rope sags by only one-fourth the distance found in (a), determine the tension in the rope. Ignore stretching of the rope. Will the rope break?

(III) You are on a pirate ship and being forced to walk the plank (Fig. 9–68). You are standing at the point marked C. The plank is nailed onto the deck at point A, and rests on the support 0.75 m away from A. The center of mass of the uniform plank is located at point B. Your mass is 65 kg and the mass of the plank is 45 kg. What is the minimum downward force the nails must exert on the plank to hold it in place?


(II) Two cords support a chandelier in the manner shown in Fig. 9–4 except that the upper cord makes an angle of 45° with the ceiling. If the cords can sustain a force of 1660 N without breaking, what is the maximum chandelier weight that can be supported?

(III) A door 2.30 m high and 1.30 m wide has a mass of 13.0 kg. A hinge 0.40 m from the top and another hinge 0.40 m from the bottom each support half the door’s weight (Fig. 9–69). Assume that the center of gravity is at the geometrical center of the door, and determine the horizontal and vertical force components exerted by each hinge on the door.
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