You are pulling your younger sister along in a small wheeled cart. You weight 65

Question

You are pulling your younger sister along in a small wheeled cart. You weight 65.0 kg and the combined mass of your sister and the cart is 35.0 kg. You are pulling the cart via a short rope which you pull horizontally. You hold one end of the rope and your sister holds the other end. If you are accelerating at a rate of 0.10 m s-2, the rope is inelastic, and the frictional force acting upon the cart is 30 N:

1) What is the tension in the rope?
2) What forces are applying to the ground in order to produce this acceleration?

(note: please answer all of the questions and explain how you got the answer. Number each answer please)

Explanation / Answer

If you make the free body diagram of both the bodies individually we will see that there is no horizontal force for the 65 kg man to provide the acceleration but actually the ground Normal Reaction force is acting at some angle (theta) whose component Ncos(theta) is giving the forward force so individually if you make the force equations you will get Ncos(theta) - T = 65*.1 (Horizontal Equilibrium) Also we have Nsin(theta) = 65*9.8 (vertical equilibrium) and also for the cart we have the equation T - (Frictional Force) = 35*.1 Here T is the tension force so on solving we get T = 3.5+30 = 33.5 N And as said before the normal reaction from the ground is giving the acceleration.

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