Question

A 5kg block is placed on a 2kg pallet which sits on the floor. The coefficients of friction between the block and the pallet are 0.6 and 0.8. The coefficients of friction between the pallet and the floor are 0.3 and 0.4. A rope is attached to the 5kg block and pulls horizontally with an unknown force Fa.

a) If the rope pulls to the right so the pallet slides along the floor, but the block does NOT slide on the pallet, what should the magnitude of Fa be so the pair accelerates at 2 m/s^2? Why? (Rope attached to the 5kg block)

b) If they accelerate at 2m/s^2 what is the magnitude of friction force that the block exerts on the pallet? Why? (Rope attached to the 5kg block)

c) What is the maximum acceleration the pallet can achieve? Why? (Rope attached to the 5kg block)

Answer 1

Value of frictions,

between block and pallet:

if block is not moving wrt pallet : µ_sM_bg = (0.8)*5*10 = 40N (but it is maximum friction, the actual value may vary between 0 to max)

if block is moving wrt pallet: µ_kM_bg = (0.6)*5*10 = 30N

Between pallet and floor:

if pallet is moving wrt ground = µ_kM_pg = (0.3)*2*10 = 6N

if pallet is not moving wrt ground = µ_sM_pg = (0.4)*2*10 = 8N (but it is maximum friction, the actual value may vary between 0 to max)

a) First of all, as we know that the pallet is moving wrt ground hence the friction on the pallet due to ground will be kinetic and its direction will be opposite to the direction of Motion (in the direction of force exerted on 5 kg block)

and we also know that the friction force between block and pallet is static (which is only responsible for acceleration of the pallet), as they both are moving with same acceleration,

applying second law for pallet,

F_bp - F_pg = M_pa

F_bp = F_pg + M_pa

F_bp = 6 + 2*2 = 10N

Now using this value and applying second law of motion on block,

F_a - F_bp = M_ba

F_a = F_bp + M_ba

F_a = 10 + 5*2 = 20N

b) as calculated above friction between block and pallet (F_bp ) is 10N , as the requirement of the friction between block and pallet is less for them to accelerate together. and it lies in the range of (0N, 40N), hence this value is possible.

c)The maximum acceleration a pallet can achieve will be when the friction between block and pallet become maximum (as that is the only main driving force for pallet), and its value can be 40N,

Now applying second law of motion on pallet,

F_bp - F_pg = M_pa

40 - 6 = 2*a

a = 34/2 = 17m/s²