Newton's laws - Stacked blocks

Problem Statement:

Two boxes of masses m1 = 10 kg and m2 = 2 kg are stacked one on top of the other (see figure). A person pulls horizontally with a force F on the lower box. Knowing that the coefficient of static friction between the two boxes is μs = 0.2, determine the maximum force that the person can exert so that the upper box does not slip off the lower one. Calculate the acceleration of the boxes for this value of force. Neglect the friction with the ground.

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Solution:

First let’s draw the forces that act on each of the boxes:

We have represented the forces that act on the two boxes on each figure. The forces acting on the lower box have been highlighted in figure (a) and those acting on the upper box in figure (b).

Note that some of the forces form action-reaction pairs: N12 is the normal force that the upper box exerts on the lower one and N21 is the normal force exerted by the lower box on the upper one. Both forces have the same magnitude but opposite directions.

FR12 is the friction force experienced by the lower box and FR21 is the friction force experienced by the upper box. Both forces constitute another action-reaction pair and as in the case of the normal forces they have the same magnitude and opposite directions.

None of the above normal forces correspond to the weight of the boxes. The weight of the lower box is the gravitational force exerted by the Earth on it, and its reaction is the gravitational force exerted by the lower box on the Earth (not represented in the figure because it is applied in the center of the Earth).

The normal N is the force that the ground exerts on the lower box and its reaction is the normal that this box exerts on the ground. It is not represented in the figure because we are interested in this problem by the motion of the two boxes, not that of the ground

A force and its reaction must be forces of the same type and are applied to different bodies.

Next we will write Newton’s second law for each of the boxes:

Box 1 (lower box)

Taking the forces highlighted in figure (a) as a reference, Newton’s second law for box 1 is:

The projection on the axes represented on the left of the figure is:

The projection of the acceleration vector is positive because the boxes move to the right (in the positive direction of the x-axis).


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Box 2 (upper box)

Taking the forces highlighted in figure (b) as a reference, Newton’s second law for box 2 is:

The projection on the axes represented on the left of the figure is:

As for block 1, the projection of the acceleration vector is positive because the boxes move to the right (in the positive direction of the x-axis).

With these three equations we can now solve the problem.

We have to calculate the maximum magnitude of F so that the two boxes move together. It means that the two boxes have the same acceleration:

In addition, the static friction force magnitude gets its maximum value (because we are calculating the maximum magnitude for force F) when:

From equation (4) we get the magnitude of the normal N21:

After substituting both expressions in equation (3) we obtain the acceleration of the masses:

And from equation (1) we obtain the maximum force magnitude so that the upper box does not slip off the lower one:

In the problem we have used g = 10 m/s2

Do not forget to include the units in the results of the problem.

When the lower box is pulled with less force than the calculated maximum, the two boxes move together. If the force is greater, the upper box will slip and the boxes will have different accelerations.

 

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