5.1 A gymnast hangs from the lower end of a rope connected to an O-ring that is

Question

5.1

A gymnast hangs from the lower end of a rope connected to an O-ring that is bolted to the ceiling. The weights of the gymnast, the rope, and the O-ring are 500 N. 100 N, and 50 N. respectively. What are the magnitudes of the tensions at both ends of the rope? SET UP First we sketch the situation (Figure 1). Then we draw three free-body diagrams, for the gymnast, the rope, and the O-ring (Figure 2). The forces acting on the gymnast are her weight (magnitude 500 N) and the upward force (magnitude T_1) exerted on her by the rope. We don't Include the downward force she exerts on the rope because it isn't a force that acts on her. We take the y axis to be directed vertically upward, the x axis horizontally. There are no x components of force; that's why we call this a one-dimensional problem. SOLVE The gymnast is motionless, so we know mat she is in equilibrium. Since we know her weight, we can use the equation. Sigma F_y = 0. to find the magnitude of the upward tension T_1 on her. This force pulls In the positive y direction. Her weight acts in the negative y director, so Its y component Is the negative of the magnitude -that is. -500 N. Thus, Sigma F_y = 0 T_1 + (- 500 N) = 0 (equilibrium of gymnast) T_1 = 500 N 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 (Figure 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_1 is 500 N. 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 T_2) exerted on its upper end by the O-ring. The equilibrium condoning Sigma F_y = 0 for the rope gives Sigma F_y = T_2 +(-100 N) + (-500 N) = 0 (equilibrium of rope) T_2 = 500 N REFLECT The tension is 100 N greater at the top of the rope (where it must support the weights of both the 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-ring as the gymnast hangs from the bottom end of the rope? Express your answer in newtons to three significant figures.

Explanation / Answer

If we consider the equilibrium of the rope at top and bottom then we have following cases:

At bottom

vertical force acting on rope = weight of gymnast = 500N

Hence, tension in rope at bottom = 500N

At top

Total vertical force acting at top of rope = weight of gymnast + weight of rope = 500+100 = 600N

Part A

Force exerted by O-ring to the bolt = total vertical force at that junction = weight of gymnast+weight of rope+weight of O-ring

= 500+100+50= 650N

Per Newton's third law, every action has equal and opposite reaction.Therefore, force exerted by bolt on O-ring = 650N

Related Questions

Navigate

• Browse (All)
• Browse 5
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.