Question

Two blocks of masses M 1 and M 2 are connected by a massless string that passes over a massless pulley as shown in the figure. M 2 , which has a mass of 13.5 kg , rests on a long ramp of angle θ = 35.5 ∘ . Friction can be ignored in this problem. Find the value of the mass M 1 for which the two blocks are in equilibrium (i.e., not accelerating).

Answer 1

The mass M1 is 7.8 kg

Explanation:

Block M1 is hanging on the string while block M2 is on the frictionless ramp.

We have to write the equations of motion for the two blocks.

- For M1, the only two forces acting on it are the force of gravity $M_1 g$ (downward) and the tension in the string T (upward). So we can write

$M_1 g - T = M_1 a$

where

$M_1$ is the mass of the block

$g=9.8 m/s^2$ is the acceleration of gravity

$a$ is the acceleration of the system

- For M2, the only two forces acting on it are the tension in the string T (acting up along the ramp) and the component of the gravity acting down along the ramp, $M_2 g sin \theta$. So the equation of motion is

$T-M_2 g sin \theta = M_2 a$

where

$M_2 = 13.5 kg$ is the mass of the 2nd block

$\theta=35.5^{\circ}$ is the angle of the ramp

In order for the two blocks to be in equilibrium, the acceleration must be zero:

$a=0$

So the two equations become:

$M_1 g - T=0\\T-M_2 g sin \theta = 0$

Isolating T from the 1st equation,

$T=M_1 g$

And substituting into the 2nd equation, we can find the value of the mass $M_1$:

$M_1 g - M_2 g sin \theta = 0\\M_1 = M_2 sin \theta = (13.5)(sin 35.5^{\circ})=7.8 kg$

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