The drawing shows three boxes of masses m_A, me and m_c (all equal to 1 kg), res

Question

The drawing shows three boxes of masses m_A, me and m_c (all equal to 1 kg), respectively, sitting on the floor of a truck. The truck accelerates up a hill with a magnitude of a_T = 5 m/s^2. The hill is inclined at an angle theta = 30degree with the horizontal. There is friction between all surfaces, but the coefficients of friction vary as follows: Between A and B |mu_s = 0.4 and mu_k = 0.2. Between B and C, mu_k = 0.5 and mu_k = 0.3. And between C and the truck, mu_s = 0.6 and mu_k = 0.4 Calculate the acceleration (magnitude and direction) of each of the blocks.

Explanation / Answer

For the block A:

Here the net downward (along the incline) force acting on the block Fa = m*g*sin + m*aT  (component of weight and psedo force)

So Fa  = 1*10*sin30 + 1*5 = 10 N

Also the normal reaction on the block will be Ra = m*g*cos = 1*10*cos30 = 8.66 N

Hence the maximum frictional force possible on the block fmax(a) = s(a) * Ra = 0.4*8.66 = 3.46 N < fa

Hence the block A will move down the incline with acceleration aA = (fa - fk(a))/m = (10 - 0.2*8.66)/1 = 8.27 m/s2 (This acceleration is with respect to the truck. Also while the block is moving then only the kinetic friction force fk = k(a) * Ra will act)

For the block B:

Normal reaction Rb = 2*m*g*cos = 17.32 N so the maximum frictional force possible fmax(b) = 0.5*17.32 = 8.66 N

Net downard force (along the incline without) Fb = m*g*sin + fk(a)  = 5 + 0.2*8.66 = 6.73 N < fmax(b)

So the block B will not move relative to block A (remains at rest in truck's frame).

For the block C:

Normal reaction Rc = 3*m*g*cos = 25.98 N

so the maximum frictional force possible fmax(c) = 0.6*25.98 = 15.59 N

Net downard force (along the incline without) Fc = m*g*sin + fmax(c) = 5 + 8.66 = 14.66 N < fmax(c)

Hence the block C will also not move with respect to the truck.

Therefore with respect to ground the acceleration of block B and C is aT = 5 m/s2 up the incline and the acceleration of block A is 8.27 - 5 = 3.27 m/s2 down the incline.

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