Question

A block of mass m1 = 3.7 kg on a smooth inclined plane of angle 41˚ is connected by a cord over a small frictionless pulley to a second block of mass m2 = 2.1 kg hanging vertically. Take the positive direction up the incline and use 9.81 m/s2 for g.
What is the acceleration of each block?

Answer 1

A block of mass $m_{1} = 3.7$ kg on a smooth inclined plane of angle $41{}^{\circ}$ is connected by a cord over a small frictionless pulley to a second block of mass $m_{2} = 2.1$ kg hanging vertically. Take the positive direction up the incline and use $g = 9.81 \, \text{m/s}^{2}$ . What is the acceleration of each block?

Solution: The force diagram is shown on the picture below:

$F_{w} = m\cdot g -$ weight force; N - support reaction normal force; $\alpha$ - angle of the plane slope;

First block pulls the system with force $F_{1} = F_{W1} \cdot \sin \alpha = m_{1} \cdot g \cdot \sin \alpha$ , which has a negative direction;

Second block pulls the system with force $F_{2} = F_{W2} = m_{2} \cdot g$ , which has a positive direction;

We will calculate the resultant force, and the sign of it will show the direction of movement:

$F = F_{2} - F_{1} = m_{2} \cdot g - m_{1} \cdot g \cdot \sin \alpha = g \cdot (m_{2} - m_{1} \cdot \sin \alpha) = 9.81 \cdot (2.1 - 3.7 \cdot \sin 41{}^{\circ}) = -3.21 \, \text{N}$ ;

Minus sign shows, that system will move down the incline; according to the second Newton's law, acceleration of blocks is: $a = F / m = F / (m_1 + m_2) = 3.21 / (2.1 + 3.7) = 0.553 \, \text{m/s}^2$ ;

Answer: Block 1 will move down the incline, and block 2 – upwards, and both of them will have the acceleration $0.553 \, \text{m/s}^2$ .