How much?

Classical Mechanics Level 2

A wedge of mass m 2 = 12 kg m_2=12\text{ kg} and angle of inclination θ = 4 5 ∘ \theta = 45^\circ is kept on a spring balance. A small block of mass m 1 = 2 kg m_1=2\text{ kg} moves along the frictionless incline of the wedge.

What is the reading ( \big( in kg/s 2 ) \text{kg/s}^2\big) of the spring balance as the small block slides down the wedge?

Ignore recoil of the wedge. Acceleration due to gravity is g = 9.8 m/s 2 . g=9.8\text{ m/s}^2.


The answer is 127.4.

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1 solution

Parth Sankhe
Dec 3, 2018

If N N is the normal reaction between the two objects,

N = m 1 g cos ⁡ θ N=m_1g\cos \theta

Force on weighing machine = m 2 g + N cos ⁡ θ m_2g + N\cos \theta

Put in the values to get the answer as 13 × 9.8 = 127.4 13×9.8=127.4 newtons.

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