A block P of mass 3.56 kg is lying on a rough inclined plane of angle θ = 0.73 radians. Block P is attached via a model string that passes over a model pulley to a second block Q of unknown mass M kg hanging vertically as shown in diagram. The coefficient of static friction between the block P and the inclined plane is µ = 0.3.

Question

A block P of mass 3.56 kg is lying on a rough inclined plane of angle θ = 0.73 radians. Block P is attached via a model string that passes over a model pulley to a second block Q of unknown mass M kg hanging vertically as shown in diagram. The coefficient of static friction between the block P and the inclined plane is µ = 0.3.

Calculate the value of the mass of block Q, in kg, that is just sufficient to initiate the sliding motion of block P up the inclined plane. Give your answer to 3 decimal places.

My working:
Resolving horizontally,
T-F-m1gsinα = 0
T = F+m1gsinα

Resolving vertically,
N-m1gcosα = 0
N = m1gcosα

F = μN
F = 0.3(m1gcosα)
F = 10.47708cosα

T = 10.47708cosα+34.9236sinα

T = m2g
m2 = (10.47708cosα+34.9236sinα)/9.81
m2 = 3.16990

My answer for block Q is 3.16990.
Shouldn't the mass of block Q be higher than block P to be sufficient to initiate the sliding motion of block P up the inclined plane?
Where have I done wrong?

Answer 1

Your calculations are correct, but there seems to be a misunderstanding in the question. When it says "the mass of block Q that is just sufficient to initiate the sliding motion of block P up the inclined plane," it means the minimum mass of block Q needed to overcome the static friction and start the motion of block P.

In this case, the static friction force between block P and the inclined plane is acting in the opposite direction of the force applied by the string (Tension force). To overcome the static friction, the tension force must be greater than or equal to the maximum static friction force.

Your calculation of the tension force T is correct, but the value you obtained for m2 is incorrect. It seems like you made an error while calculating it.

Let's go through the calculations again:

1. Resolve horizontally:
T - F - m1gsinθ = 0
T = F + m1gsinθ

2. Resolve vertically:
N - m1gcosθ = 0
N = m1gcosθ

3. Calculate the maximum static friction force:
F_max = μN
F_max = μ(m1gcosθ)

4. Substitute the value of F_max into the equation for T:
T = F_max + m1gsinθ
T = μ(m1gcosθ) + m1gsinθ

5. Now, solve for m2 (mass of block Q):
T = m2g
m2 = T/g
m2 = (μ(m1gcosθ) + m1gsinθ)/g

Now, substitute the given values of μ, m1, θ, and g to solve for m2.