A box with mass 20 kg rests on a rough surface with coefficient of friction 0.8. Another box with mass 1 kg is placed on this and the coefficient of friction is 0.1. If the 20 kg box is pulled with...

A box with mass 20 kg rests on a rough surface with coefficient of friction 0.8. Another box with mass 1 kg is placed on this and the coefficient of friction is 0.1. If the 20 kg box is pulled with a force 120 N, what happens to the 1 kg box.

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

A box with mass 20 kg rests on a rough surface, the coefficient of friction between the box and the surface it is placed on is 0.8. If the box is moved, there is a resistive force equal to `F = N*mu` , in a direction opposite to that in which the box is pulled. Here, N = 20*9.8 = 196 N and `mu` = 0.8; the resistive force is 156.8 N

Another box with mass 1 kg is placed on the box with mass 20 kg and the coefficient of friction between the surface of the two boxes where they touch each other is 0.1.

A force of 120 N is applied on the 20 kg box. As the resistive force is 156.8 N, the force being applied is not enough to overcome friction and the box does not move. The box with mass 1 kg placed on the heavier box also does not move.

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valentin68's profile pic

valentin68 | College Teacher | (Level 3) Associate Educator

Posted on

The answer is correct but it remains just an observation to be made here about the resistive force. The maximum resistive force from the large box is 156.8 N and it happens just on the threshold of box moving. If the applied force is just 120 N, like in this case, the resistive force is only equal to the applied force, that is 120 N. When the applied force increases, the resistive force also increases until its maximum value (otherwise the third principle will not ve valid in this case, or said a bit different the sum of net forces on the horizontal will not be zero, and thus a backward acceleration will happen).

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