(From book : College Physics 8th eddition volume one) -(authors- Young and Gelle
ID: 1739905 • Letter: #
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(From book : College Physics 8th eddition volume one) -(authors- Young and Geller) Chapter 5 question # 30 An 80n box initially at rest is pulled by a horizontal rope ona horizontal table. The coefficients of kenitic and static frictionbetween the box and the table are 1/4 and 1/2, respectively. Whatis the friction force on this box if pull is (a) 0 N, (b) 25N, (c)39N, (d) 41N, (e) 150N? I saw the answer posted and I still dont understand how to getthe correct answer. I do not understatnd how to find frictionalforce. (From book : College Physics 8th eddition volume one) -(authors- Young and Geller) Chapter 5 question # 30 An 80n box initially at rest is pulled by a horizontal rope ona horizontal table. The coefficients of kenitic and static frictionbetween the box and the table are 1/4 and 1/2, respectively. Whatis the friction force on this box if pull is (a) 0 N, (b) 25N, (c)39N, (d) 41N, (e) 150N? I saw the answer posted and I still dont understand how to getthe correct answer. I do not understatnd how to find frictionalforce.Explanation / Answer
The force due to friction is related by the equation: Ff = *Fn Fn is what is called the 'normal force' or the forcethat pushes on an object from the surface it'scontacting. For instance, in the case of a box lying on a perfecthorizontal table, the normal force will be equal in magnitude, butopposite in direction of the force due to gravity, or"weight". If the normal force exerted on the box wasless, the table would break because it cannot support theweight. If the box were on an inclined plane, the normalforce would be some proportion of the weight depending on theangle of incline. So in this example: The force due to gravity on the block, Fg, is given as80N. So we know the normal force will also be 80N but actingin the opposite direction. If we assume that motion to theright is positive and motion upward is positive, Fn will bepositive. Static friction can be thought of as a limit. Once theapplied force exceeds this limit, motion will start and we can thenuse the coeffecient of kinetic friction which is generallylower. For our problem this limit is: Ff = usFn = .5 * 80N = 40N, or -40N since it actsopposite our motion. For part a, we are told the horizontal force on the blockis 0N, in other words it is not moving, so it isstatic. The force due to friction is 0N. For part b we are told there is an applied force now of25N. Does the block move? Well we saw inthe beginning that the applied force must be greater than40N for the block to start moving. Since no motion willoccur until we apply a force greater than 40N the friction forcewill just match our applied force so the block remainsstationary. Therefore with an applied force of 25N, the friction force is-25N. For part c the force is now 39N. Is the blockmoving? Nope. 39N<40N. So, the friction forceis now -39N. For part d the applied force is now 41N, so the block canstart to move. Since the block is moving now, we can use thecoefficiend of kinetic friction, k. In thisexample that is .25. Ff = kFn = .25 * 80N = 20N, or -20N becauseit acts against our motion. So the force due to friction is now 20N with an applied forceof 41N. For part e, the applied force is 150N. What is thefriction force? Well now that the block is moving it doesn'tmatter how much force we apply - the friction force will always be-20N. In summary: a) 0N b) -25N c) -39N d) -20N e) -20N -------------------------------------------------------------------------------------------------------------------------- I just saw that you said you had the answers. If myanswers differ I would like to know what they came up with! Thanks!Related Questions
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