Question

A block of mass M rests on a block of mass M1 which is on a tabletop. A light string passes over a frictionless peg and connects the blocks. The coefficient of kinetic friction between the blocks and between M1 and the tabletop is the same. A force F pulls the upper block to the left and the lower block to the right. The blocks are moving at a constant speed.

Determine the mass of the upper block. (Express your answer to three significant figures.)

Answer 1

M = F/3μ g - M₁/3

Step-by-step explanation:

To solve this exercise we must use the equilibrium conditions translations

∑ F = 0

In the attachment we can see a free body diagram of each block

Block M (upper)

X axis

fr₁ + F₂ -F = 0

F = fr₁ + F₂ (1)

axis

N₁-W = 0

N₁ = Mg

the friction force has the formula

fr₁ = μ N₁

F = μ Mg + F₂

bottom block

X axis

F₂ - fr₁ - fr₂ = 0

F₂ = fr₁ + fr₂

Y axis

N - W₁ -W = 0

N = g (M + M₁)

we substitute

F₂ = μ Mg + μ (M + M1) g

F₂ = μ g (2M + M₁)

we substitute in 1

F = μ M g + μ g (2M + M₁)

F = μ g (3M + M₁)

we look for mass M

M = (F - μ g M₁)/ 3μ g

M = F/3μ g - M₁/3

the exercise does not have numerical data