Question

Two blocks of mass 10 kg and 4 kg are placed over the surface of a lift accelerating upwards with a=2m/s^2. Find contact force between the blocks

Answer 1

Answer:

48N

Explanation:

Mass of block one = m1 = 10 kg (Given)

Mass of block one = m2 = 4 kg (Given)

Acceleration of the lift = 2m/s² (Given)

Let the contact force between the blocks = F.

A pseudo acceleration is always identical to the spectral acceleration. However, in the case of damping other than zero the two are slightly different. Thus, according to the concept of pseudo acceleration -

Thus, from the block of mass = m2 at equilibrium,  

F = 4 × 10 + 4 × 2

F = 40 + 8   

= 48

Therefore, the contacting force between the blocks is 48N.

Answer 2

Answer:

Explanation:

Mass of block one = m1 = 10 kg (Given)

Mass of block one = m2 = 4 kg (Given)

Acceleration of the lift = 2m/s² (Given)

Let the contact force between the blocks = F.

A pseudo acceleration is always identical to the spectral acceleration. However, in the case of damping other than zero the two are slightly different. Thus, according to the concept of pseudo acceleration -

Thus, from the block of mass = m2 at equilibrium,

F = 4 × 10 + 4 × 2

F = 40 + 8

= 48

Therefore, the contacting force between the blocks is 48N.