Stack them up!

Classical Mechanics Level 4

Two smooth cylindrical bars weighing $20\sqrt { 3 } N$ each lie next to each other in contact. A similar third bar is placed over the two bars as shown in the figure.

To keep the bottom two cylinders in contact equal and opposite horizontal forces are applied at the centers of both the bottom cylinders.

If the friction is neglected, what is the minimum horizontal force on one of the cylinders required to keep them in contact?

The answer is 10.

1 solution

Nanayaranaraknas Vahdam
Sep 17, 2014

This is just a shortened solution.

Basically, draw the free body diagram of the centre of mass's of each cylinder. Let the normal reaction from one of the bottom cylinders on the top one be $N$ . Thus, the horizontal components of the two normals cancel each other, and the vertical components cancel the weight of $20\sqrt{3}$ . By symmetry, we can say that each normal must balance $10\sqrt{3}$ .

Thus, $N\sin \theta=10\sqrt{3}$

Since the centre of masses form an equilateral triangle, $\theta=60$

Therefore, $N=20$

The action for $N$ is also $20$ newtons. Thus, the horizontal component, which must be countered by the applied force, is $20\cos 60=\boxed{10}$ newtons.

Stack them up!

The answer is 10.

1 solution

0 pending reports