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Problem 60 Hard Difficulty

The drawing shows a circus clown who weighs 890 $\mathrm{N}$ . The coefficient of static friction between the clown's feet and the ground is 0.53 . He pulls vertically downward on a rope that passes around three pulleys and is tied around his feet. What is the minimum pulling force that the clown must exert to yank his feet out from under himself?


310 $\mathrm{N}$


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Video Transcript

in this question. We have the following forces in action. So we have the weight force off the club. We have the normal force that the ground is exerting on the club. We have the force that the clown is exerting on the rope, and then we have the group answering that force. F we've attention t. And then that tension t is transferred to his feet. Moreover, legs Hauser, the frictional force that tries to keep the clowning place. Then we have to calculate what is the minimum value off this force F that is neccessary to make his feet move. So what is the minimum value off F such that these force can be bigger down the frictional force? In order to do that, we have to apply Newton's second law, and I think that we will need to apply it in these access the vertical one, which I'll call why, and l'm these access, which I will call X the horizontal axis. We begin by applying the Newton's second law toe the vertical axis, then on the vertical access. Newton's second law tells us that the Net force is close to the mask off the clown times acceleration off this clown acceleration is it goes to zero because he is standing still in the round. Then the net force on the Y direction is it goes to zero. But the net force on the wider action is composed by three forces T Normal Force and the Weight Force. These are the forces that are acting on the cloud. The Force f is a force. Produce it by the clown show. It will not enter in this equation. In this equation, we will only use the forces that are acting on the clown. Then we have that the tension force plus the normal force minors. The wait for season goes to zero. Then we have that falling relation. T plus N is equal to the weight force and the weight force is equals to 809 noodles. Then we knew relation between the tension force on the normal force. Now let as do the same thing. But for the X axis on the horizontal access the net force use Our show equals to the mass off the clown times his acceleration again. He is a standing still. He's not moving yet. Then these acceleration is close to zero. So the net force is It goes to zero on the horizontal direction. But on the direction. The net force is composed by true forces the tension force on the fictional force. Then the frictional force which is pointing to the positive direction off my horizontal access. We will have a positive sign and then we have miners, that tension force and this is the cost of zero. Now we can so for the tension. So the tension is he goes to definitional force. Now what is the maximum value off the frictional force? So we want the limit situation. So for the limiting situation, we have to complete what is the maximum value off the frictional force? And we know that the maximum value off the fictional force is equals to the static frictional coefficient, times the normal force. Then in order for the clown be almost moving, we have to have attention. That is equal studio you should know, started coefficient times the normal force. And then we have to complete what is the value off the normal force. And to do that we can go back to this equation and Soviet for the normal force by doing that, we got the following the normal force. Is he close to 819th minus detention force. Then we can use these result in these other equation and Soviet for detention. So attention force is because to the static friction aquisition times the normal force which is 890 my news attention force then teeth is equals to 890 times, um us minus us times t Then we can send this term to the other side to get teeth plus some us time. Steep cheesy goes to 890 times me west. So we factor us detention on the left hand side to get teeth times one plus mute us is equals to 890 kind to us. Finally, potential is it goes to 890 times me west, divided by one plus from us. But I'm us is equal to 0.53. Then the tension force is equals to 890 times 0.53 divided by one plus 0.50 tree. These results in attention off approximately 310 Newtons. So this is the tension force. But according to Newton's Turtle, the tension force is a reaction force, and it's reacting to the force that is applied by the clone. Therefore, the force F should be equals 2 310 Newtons in the limiting situation, so have must be bigger than 310 new terms in order for the clown to be able to move his feet.