A box of bananas weighing 36.5 N rests on a horizontal surface. The coefficient

Question

A box of bananas weighing 36.5 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.36, and the coefficient of kinetic friction is0.21.

(b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.1 N to the box and the box is initially at rest?
N
(c) What minimum horizontal force must the monkey apply to start the box in motion?
N
(d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started?
N
(e) If the monkey applies a horizontal force of 19.0 N, what is the magnitude of the friction force and what is the box's acceleration

please someone answer with correct solutions ?

Explanation / Answer

Hi,

For this case we should first see what forces are involved in this problem:

For the x axis: F - f = ma ; where F is the horiziontal force, a is the acceleration of the box, m is the mass of the box and f is the friction force.

At this point we must define f and mass.

m = W/g ; where W is the weight of the box and g is the acceleration due to gravity (which has a value of 9.8 m/s2 ).

f = N ; where N is the normal force and is the coefficient of friction.

For the y axis: N = W; so thanks to this we have the value of the friction force, because N is known now.

So: F - W = W(a/g) ::::::::: a = g*(F/W - ) (1)

(b) For the first question we must see if that force is enough to move the box in the first place. So we calculate the friction force using the static coefficient of friction:

fs = sW = (0.36)*(36.5 N) = 13.14 N; this is the maximum value of the friction force

As this force is bigger than the force applied by the monkey the box will not move, and so the friction force of the box must be equal to the force applied by the monkey.

f = 6.1 N; in fact the force should be a little higher (but just a little)

(c) The minimum force the monkey must apply to start the box in motion should be equal to the maximum value of the static friction force, which has been already calculated.

F = 13.14 N

(d) If the box must remain in movement but at constant velocity its acceleration must be zero, so the horizontal force should be equal to the kinetic friction force.

F = kW = (0.21)*(36.5 N) = 7.67 N

So the force needed to mantain the box in movement at constant velocity is:

F = 7.67 N

(e) For this case we used the equation (1) to calculated the acceleration. As the box is already in movement we use the coefficcient of kinetic friction:

a = 9.8 m/s2 ( 19/36.5 - 0.21 ) = 3.04 m/s2

I hope it helps.

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