Question

The coefficient of static friction between mass mA

Answer 1

Answer:

(a) 5.0 kg

(b) 10 kg

Explanation:

Draw a free body diagram for each block.  There are 4 forces on block A:

Weight force mAg pulling down,

Normal force N pushing up,

Tension force T pulling right,

and friction force Nμ pushing left.

There are 2 forces on block B:

Weight force mBg pulling down,

and tension force T pulling up.

Whether the system is just starting to move, or moving at constant speed, the acceleration is 0.

Sum of forces on B in the -y direction:

∑F = ma

mBg − T = 0

mBg = T

Sum of forces on A in the +y direction:

∑F = ma

N − mAg = 0

N = mAg

Sum of forces on A in the +x direction:

∑F = ma

T − Nμ = 0

T = Nμ

Substitute:

mBg = mAg μ

mA = mB / μ

(a) When the system is just starting to move, μ = 0.40.

mA = 2.0 kg / 0.40

mA = 5.0 kg

(b) When the system is moving at constant speed, μ = 0.20.

mA = 2.0 kg / 0.20

mA = 10 kg