Question

the mass of block A is 80 kg, the mass of block b is 20 kg, and A is connected to B with the cable and pulley system shown. the pulleys rotate freely, and the cable and pulleys have negligible mass. the coefficient of static fricion between A and the horizontal surface is Ms=0.4, and the coefficient of kinetic friction is Mk=0.3.

a) initially, block A is being held stationary and the system is at rest in static equilibrium . Find tension in the cable(answer: 196.2 N)

b) the system is released from rest. show the colfficient of static friection is not large enough to keep block A from sliding.

c) Find the acceleration of block A and the tension in the cable when block A starts sliding ( answers: aA=0.981 m/s^2 and 156.96 N)

Answer 1

tension in the cable must support block b from falling

T = mg = 20*9.81 = 196.2 N

can you provide the figure please?

am giving you the procedure for now

static friciton = .4 * Normal reaction(N)

Force on 80 kg is 2T = 2*196.2 = 392.4 N

static friction = .4*80*9.81 =313.92

static friction is less than the force

so not large enough

c)

when it starts sliding

friction = .3*Normal reaction

for 80 kg mass , let acceleration be a

2T- .3*80*9.81 = 80*a

for 20 kg mass, acceleration is 2a (for 1 unit movement of 80 kg 2 units of 20 kg will move down)

20*9.81 - T = 20*(2a)

solving we get

a = .981 m/s^2

T = 156.96 N