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A woman is balancing on a high wire, which is tightly strung, as shown in Fig. 9–45. The tension in the wire is

(a) about half the woman’s weight.

(b) about twice the woman’s weight.

(c) about equal to the woman’s weight.

(d) much less than the woman’s weight.

(e) much more than the woman’s weight.

Short Answer

Expert verified

The correct option is (e).

Step by step solution

01

Concepts

At equilibrium, the net force is zero. For this problem, find the condition for vertical forces.

02

Explanation

Let m be the mass of the woman and T be the tension in the wire.

At equilibrium for the vertical forces,

\(\begin{aligned}{c}T\sin \theta + T\sin \theta = mg\\2T\sin \theta = mg\\\sin \theta = \frac{{mg}}{{2T}}\end{aligned}\).

You can observe that the angle of the rope with the horizontal is very small and less than \({30^ \circ }\). Then,

\(\begin{aligned}{c}\sin \theta < \sin {30^ \circ }\\\frac{{mg}}{{2T}} < \frac{1}{2}\\T > mg\end{aligned}\).

Therefore, the tension in the wire is more than the woman’s weight.

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Most popular questions from this chapter

A shop sign weighing 215 N hangs from the end of a uniform 155-N beam, as shown in Fig. 9–58. Find the tension in the supporting wire (at 35.0°) and the horizontal and vertical forces exerted by the hinge on the beam at the wall. [Hint: First, draw a free-body diagram

A 10.0 N weight is suspended by two cords, as shown in Fig. 9–44. What can you say about the tension in the two cords?

(a) The tension in both cords is 5.0 N.

(b) The tension in both cords is equal but not 5.0 N.

(c) The tension in cord A is greater than that in cord B.

(d) The tension in cord B is greater than that in cord A.

(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).

(III) A steel cable is to support an elevator whose total (loaded) mass is not to exceed 3100 kg. If the maximum acceleration of the elevator is \({\bf{1}}{\bf{.8}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\), calculate the diameter of the cable required. Assume a safety factor of 8.0.


Question:The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.2 m above the ground, how steep a slope can the truck be parked on without tipping over (Fig. 9–81)?



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