A boy of mass m pulls (horizontally) a sled of mass M . the coefficient of frict
ID: 1685956 • Letter: A
Question
A boy of mass m pulls (horizontally) a sled of mass M. the coefficient of friction between sled and snow is µ. a) draw a diagram showing all forces acting on the boy and on the sled. b) find the horizontal and vertical components of each force at a moment when boy and sled each have an acceleration a.c) if the coefficient of static friction between the boy's feet and the ground is µs, what is the maximum acceleration he can give to himself and the sled, assuming traction to be the limiting factor? A boy of mass m pulls (horizontally) a sled of mass M. the coefficient of friction between sled and snow is µ. a) draw a diagram showing all forces acting on the boy and on the sled. b) find the horizontal and vertical components of each force at a moment when boy and sled each have an acceleration a.
c) if the coefficient of static friction between the boy's feet and the ground is µs, what is the maximum acceleration he can give to himself and the sled, assuming traction to be the limiting factor?
Explanation / Answer
We can consider the motion of the boy as the motion of a toboggan.The toboggan is no longer in equilibrium,but accelerates as it slides down the hill.Hence we must use Newton's second law,SF = m * a,in its component form.Our target variable is the downhill acceleration. The toboggan's y-component of acceleration ay is still zero,but the x-component ax is not. It's convenient to express the weight as w = mg.Then,using Newton's second law for the x- and y-components of force,we have SFx = mg * sina + (-fk) = m * ax and SFy = n + (-mg * cosa) = 0 From the second equation,we get n = mg * cosa and fk = µk * n = µk * mg * cosa We substitute this back into the x-component of equation.The result is mg * sina + (-µk * mg * cosa) = m * ax or ax = g * (sina - µk * cosa) Here are some special cases we can check.First,if the hill is vertical,a = 90 o;then sina = 1,cosa = 0,and ax = g.This is free fall,just what we would expect.Second,on a hill at angle a with no friction,µk = 0.Then ax = g * sina.Related Questions
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