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?
1 answer