Question

A block (mass m1) lying on a frictionless inclined plane is connected to a mass (mass m2) by a massless cord passing over a pulley, as shown in Figure below. (a) Determine a formula for the acceleration of the system of the two blocks in terms of m1, m2, theta, and g. (b) What conditions apply to masses andm1 and m2 for the acceleration to be in one direction (say, m1 down the plane), or in the opposite direction? Show all work, please!

Answer 1

Answer:

a) System aceleration:

• $a=\frac{g(m_2-m_1sin(\theta))}{m_1+m_2}$

b) Direction of movement:

The block $m_1$ down the plane when the acceleration is negative. This occur when:

$m_2-m_1sin(\theta)$

The block $m_2$ up the plane when the acceleration is positive. This occur when:

$m_2-m_1sin(\theta)>0$

Explanation:

For the block $m_1$ the move direction is parallel (||) to the plane

$\sum F_{||}=m_1a=T-sin(\theta)mg$  (1)

For the block $m_2$  the move direction is vertical (y)

$\sum F_y=m_2a=m_2g-T$  (2)

Both blocks are connected with the same cable, therefore, the tension on the cable and the acceleration is the same for both.

Solving the equation 2 for T:

$T=m_2g-m_2a$   (3)

replacing (3) in the equation (1)

$m_2g-m_2a- m_1gsin(\theta)=m_1a$  (4)

solving (4) for a:

$a=\frac{g(m_2-m_1sin(\theta))}{m_1+m_2}$