Q3.6 Normal contact force (S)

Two blocks of masses m1 and m2 are moved upwards against gravity by the action of an upward force of magnitude FA. What is the normal contact force between the two masses?
Graphic - No title - twoblocks
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Hint

This is like the example analysed in S2.6. But now you’ve got gravity to contend with too.
 

Solution

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Solution

Let a denote the upward acceleration of the system. First we apply Newton’s 2nd law to the system comprising both blocks. The forces acting on this system (draw a box around it and think of the influences from outside the box) are

  1. the applied force, which has magnitude FA and points up;
  2. the total weight (gravitational force) which has magnitude (m1+m2)g and points down.

Then

FA-(m1+m2)g=(m1+m2)a
The left hand side gives the net force acting in the upward direction; the rhs gives the mass times the acceleration in this direction. Rearranging gives
FA=(m1+m2)(g+a)
Now consider the system comprising the upper mass alone. The forces acting on this system are (draw the box!)

  1. the normal contact force exerted on it by the lower mass; this has unknown magnitude FN and points up;
  2. its weight (the gravitational force it experiences) which has magnitude m2g and points down.

Now Newton’s 2nd law gives

\[ F_N-m_2g =m_2a \hspace{0.5cm}\mbox{\rm or}\hspace{0.5cm} F_N=m_2(g+a) \]

Note that we are right to use the same symbol for the acceleration in both of the above equations. As long as the two blocks move together they have the same acceleration.

Now dividing our result for FN by our result fo FA gives

\[ \frac{F_N}{F_A} = \frac{m_2}{m_1+m_2} \]

It is interesting to note that this result is just the same as the one we found (cf S2.6) in the case in which the force is applied horizontally.