Question

Two blocks with rough surfaces are stacked on top of a slippery horizontal surface. The top block has a mass of m, and the bottom block a mass of 2m. A rope of negligible mass is attached to the bottom block and the other end is hung over a pulley with negligible

friction and mass, where it is attached to a hanging block of mass 3m. The friction force between the two blocks keeps the top block

from sliding off the bottom block, and the hanging block is released as the system begins at rest. Find the contact forces, the friction force, the tension force, and the distance the hanging weight drops in a time t. Your answers should all be in terms of m, g, and t.

Please include a free body diagram for each of the blocks.

Answer 1

Answer:

Please refer to the attached schematic for free-body diagrams of the blocks.

The contact forces N1 = mg, N2 = 3mg.

The friction force fs = mg/2.

The tension force T = 3mg/2.

The distance y = (gt^2)/4

Explanation:

Starting with the third block which is declining a time t. Newton's Second Law implies that

$F = (3m)a\\3mg - T = 3ma\\T = 3mg - 3ma$

There are two unknowns in one equation: T and a. Therefore, we need to find more equations to solve these unknown variables.

Let's look at free-body diagram 1:

Newton's Second Law:

$F = ma\\f_s = ma$

That is another equation with one more unknown: fs.

Free-body diagram 2:

$F = (2m)a\\T - f_s = 2ma$

That is our third equation. We now have three equations with three unknowns. Combining equation 2 and 3 gives:

$T - ma = 2ma\\T = 3ma$

Plugging this equation into the first equation gives

$T = 3mg - 3ma\\3ma = 3mg - 3ma\\3mg = 6ma\\a = g/2$

Following this, we can find the other unknowns:

$T = 3ma = 3m(g/2) = \frac{3mg}{2}$

$f_s = ma = mg/2$

The contact forces are N1 and N2.

$N_1 = mg\\N_2 = 2mg + N1 = 3mg$

Finally, using the acceleration and equation of kinematics, we can find the distance the hanging weight drops in a time t.

$y - y_0 = v_0t + \frac{1}{2}at^2\\y = \frac{1}{2}(\frac{g}{2})t^2 = \frac{gt^2}{4}$