86054339 5.1 si 5.1 SOLUTION from the lower end of a rope SET UP First we sketch

Question

86054339 5.1 si 5.1 SOLUTION from the lower end of a rope SET UP First we sketch the situation (Eigure 1). Then we draw three free-body diagrams, for the gymnast the rope, and the O-ring and the upward force ( g (Figure 2). The forces acting on the gymnast are her weight (magnitude 500 N) The weights of the gymnast, the rope, and the O- ring are 500 N, 100 N, and 50 N, respectively. What are the ends of the rope? Ti) exerted on her by the rope. We don't include the force she exerts on the rope because it isn't a force that acts on her. We take the y axis to be directed vercally upward, the z axis horizontally. There are no z components of force; that's why we call this a one- of the tensions at both can use the equation. Fv =0. to find the magritude of the upward tensionT, on her This force pulls in Her weight acts in the negative y direction, so its y component is the negative of the Ti (-500 N) 0 (equilibrium of gymnast) T,=500N The two forces acting on the gymnast are not an action-reaction pair because they act on the same object Next we need to consider the forces acting on the rope (Eiqure 2). Newton's third law tells us that the gymnast exerts a force on the rope that is equal and opposite to the force it exerts on her. In other words, she pulls down on the rope with a force whose magnitude T is 500 N. :of2 As you probably expect, this force equals the gymnast's weight. The other forces on the rope are its own weight (magnitude 100 N) and the upward force (magnitude Ta) exerted on its upper end by the O-ring. The equilibrium, condorEF-0 for the rope gives E, T) + (-100 N) + (-500 N) = 0 (equilibrium of rope) 600 N = = REFLECT The tension is 100 N greater at the top of the rope (where it must rope and the gymnast) than at the bottom (where it supports only the gymnast). What is the magnitude of the force exerted by the bolt on the O-ting as the gymnast hangs from the

Explanation / Answer

If we draw the free body diagram of the O-ring then we notice that force acting on it are

(1) Since O-ring exerts a force T2 on the rope in the upward direction,from Newton's third law rope also exerts a same force T2 on the O-ring in the downward direction.

(2) The weight of the O- ring acting downward (W)

(3) The force exerted by the bolt (F) on the O-ring in the upward direction.

Since the O-ring is at rest the summation of all vertical forces on O-ring must be zero

F - T2 - W =0

F = T2 + W

F = 600 + 50

F = 650 N

Hence the force exterted by the bolt on the ring is 650 N

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