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Chapter 9: Q17Q (page 230)

Is the Young’s modulus for a bungee cord smaller or larger than that for an ordinary rope?

Short Answer

Expert verified

The Young’s modulus for a bungee cord is smaller than that of an ordinary rope.

Step by step solution

01

Understanding Young’s modulus

Young’s modulus is one of the essential properties of an element that estimates the stiffness and identifies the ductility or brittleness of the element.

02

Evaluating the change in Young’s modulus with ordinary rope and bungee cord

The relation of Young’s modulus is given by:

\(E = \left( {\frac{F}{A}} \right)\left( {\frac{L}{{\Delta L}}} \right)\)

Here, Fis the force, Ais the area, Lis the length of the cord, and\(\Delta L\)is the change in the length of the cord.

In the above relation, it is observed that Young’s modulus relies on the change in the length of the particular chord. This indicates that the value of Young’s modulus reduces with an increase in change in length.

The variation in the length of a bungee cord is much larger than an ordinary rope. So, Young’s modulus for a bungee cord will be smaller.

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Most popular questions from this chapter

A 23.0-kg backpack is suspended midway between two trees by a light cord as in Fig. 9–51. A bear grabs the backpack and pulls vertically downward with a constant force, so that each section of cord makes an angle of 27° below the horizontal. Initially, without the bear pulling, the angle was 15°; the tension in the cord with the bear pulling is double what it was when he was not. Calculate the force the bear is exerting on the backpack.

(II) The subterranean tension ring that exerts the balancing horizontal force on the abutments for the dome in Fig. 9–34 is 36-sided, so each segment makes a 10° angle with the adjacent one (Fig. 9–77). Calculate the tension F that must exist in each segment so that the required force of\(4.2 \times {10^5}\;{\rm{N}}\)can be exerted at each corner (Example 9–13).

Explain why touching your toes while you are seated on the floor with outstretched legs produces less stress on the lower spinal column than when touching your toes from a standing position. Use a diagram?

(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?

A 2.0-m-high box with a 1.0-m-square base is moved across a rough floor as in Fig. 9–89. The uniform box weighs 250 N and has a coefficient of static friction with the floor of 0.60. What minimum force must be exerted on the box to make it slide? What is the maximum height h above the floor that this force can be applied without tipping the box over? Note that as the box tips, the normal force and the friction force will act at the lowest corner.
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