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two blocks are connected via a string draped over frictionless pulley. assume th

ID: 1262742 • Letter: T

Question

two blocks are connected via a string draped over frictionless pulley. assume that the string and the pulley have negligible mass.
A) what is the minimum value of the coefficient of static friction that would keep the blocks from moving ?
B) what is the acceleration on the moving blocks when the coefficient of kinetic friction UK is .070?
show your work .


two blocks are connected via a string draped over frictionless pulley. assume that the string and the pulley have negligible mass. A) what is the minimum value of the coefficient of static friction that would keep the blocks from moving ? B) what is the acceleration on the moving blocks when the coefficient of kinetic friction UK is .070? show your work .

Explanation / Answer

a) Both blocks are at rest , so a = 0

on vertically hanging block in vertical direction,

Fnet = T - mg = 0

T = mg = 2g


Block on the incline :

Perpendicular to the incline ,

Fnet = N - mgcos30 = 0

N = 5gcos30


along the incline,

Fnet = T + friction - mgsin30 = 0

friction = u.N = u5gcos30

putting the values,

2g + u5gcos30   = 5gsin30

u = (5sin30 - 2) / 5cos30 = 0.115 .....Ans (A)


B) 5kg block will move along downwards. and 2kg will move upwards.

on vertically hanging block in vertical direction,

Fnet = T - mg = ma

T - 2g = 2a .....(i)


Block on the incline :

Perpendicular to the incline ,

Fnet = N - mgcos30 = 0

N = 5gcos30


along the incline,

Fnet = 5gsin30 - T - f = 5a

friction = u.N = 0.070 *5gcos30 = 0.35gcos30

5gsin30 - 0.35gcos30 - T = 5a

2.2g - T = 5a .....(ii)

adding (i) and (ii) ,

2.2g - 2g = (5 +2)a

a = 0.276 m/s2