An inventive child named Nick wants to reach an apple in a tree without climbing

Question

An inventive child named Nick wants to reach an apple in a tree without climbing the tree. Sitting in a chair connected to a rope that passes over a frictionless pulley (see figure below), Nick pulls on the loose end of the rope with such a force that the spring scale reads 280 N. Nick's true weight is 290 N, and the chair weighs 160 N. Nick's feet are not touching the ground. Find Nick's acceleration, using upward as positive.
Find the magnitude of the force Nick exerts on the chair. Instead Nick hands the rope with the scale to his friend Barney, who stands on the ground. Barney pulls on the rope so that the spring scale again reads 280 N. What is Nick's acceleration now, again using upward as positive.

Explanation / Answer

You should first draw a picture of the problem. I can't draw it here, but I envision the boy sitting in the chair, a rope extending upward from the chair over a pulley (on a tree branch overhead), and extending back down to the boys hand. The boy then pulls downward on the end of the rope to lift himself and the chair up to the apple. It's important to draw this to properly draw a force diagram. You would see that the forces pulling downward are gravity acting on the boy and the chair. But you would also see that there are two forces pulling up on the boy and the chair - these forces are the tensions on both ends of the rope hanging from the pulley. Since the tension in the rope is the same everywhere, both of these tension are equal to the 293N pull of the child. Now you can set up and equation based on Newton's 2nd Law: F = ma = sum of forces acting on object The mass of the boy is 431N/(9.8m/s2) = 44.0 kg. The mass of the chair is 137N/(9.8m/s2) = 14.0 kg So the equation is: (440kg + 14.0 kg)*a = 2*293N - 431N - 137N = 18N Now solve for a: a = 18N/(440kg + 14.0kg) = 0.040 m/s2 (upwards) For part b). the force Chris exerts on the chair should just be its mass x acceleration: F(on chair) = (14.0kg)*(0.04m/s2) = 0.55N This one is a little bit tricky in how you interpret it. This is the force being exerted on the chair (technically by the two ends of the rope together). Since this is a pulley system where the rope is exerting tension in two spots simultaneously, Chris actually only needs to exert half as much force to accomplish the same work on the chair (0.275N) - that's why pulleys are so useful (I hope you can understand that description) I hope this helps.

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