Question

Displayed to the right are two carts connected by a cord that passes over a small frictionless pulley. Each cart rolls freely with negligible friction. Consider the mass of the cart to be m1 and the angle the first cart sits on to be θ1, while the mass of the second cart as m2 and the angle it sits on to be θ2.

Answer 1

The acceleration is given by the tension in the rope and the component

of the component of the weight acting along the plane.

Response:

$The \ expression \ for \ the \ acceleration, \ a \ is\ \underline{a = \dfrac{m_2 \cdot g \cdot sin \left(\theta_2 \right) - m_1 \cdot g \cdot sin \left(\theta_1 \right) }{m_1 + m_2}}$

How can the expression for the acceleration be derived?

A possible question is the expression for the acceleration, a in terms of the mass, m₁, m₂, and the angles θ₁ and θ₂.

The drawing representing the situation, obtained from a similar question and created with MS Visio, is attached

From the drawing, we have;

By equilibrium of forces, the tension in the cable, T, is given as follows;

For the cart, m₁; T = m₁·a + m₁·g·sin(θ₁)

For the cart, m₂; T = m₂·g·sin(θ₂) - m₂·a

Where;

a = The acceleration that the two carts experience

Equating both values of T gives;

m₁·a + m₁·g·sin(θ₁) = m₂·g·sin(θ₂) - m₂·a

m₁·a + m₂·a =  m₁·g·sin(θ₁) - m₂·g·sin(θ₂)

The

$a = \mathbf{\dfrac{m_2 \cdot g \cdot sin \left(\theta_2 \right) - m_1 \cdot g \cdot sin \left(\theta_1 \right) }{m_1 + m_2}}$

The acceleration, a, that the two carts experience in terms of m₁, m₂, θ₁, θ₂,

is therefore;

• $a = \underline{\dfrac{m_2 \cdot g \cdot sin \left(\theta_2 \right) - m_1 \cdot g \cdot sin \left(\theta_1 \right) }{m_1 + m_2}}$

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