Question

You apply a coefficient of static friction is 0.56. (a) What is the force (in N) in the knee joint of a person who 56.0 kg of her mass on that: knee if the it is such can cause further damage and pain. Assume that the c of kinetic friction in the joint is 0.015

Answer 1

048)

m = mass of refrigerator = 61 kg

N = normal force on the refrigerator

$\mu$_s= Coefficient of static friction = 0.56

f_smax = static frictional force that can act on the refrigerator

the normal force on the refrigerator by the ground in upward direction balances its weight in down direction, hence the force equation in vertical direction is given as

N - mg = 0

N = mg

N = 61 x 9.8 = 597.8 N

maximum static frictional force that can act on the refrigerator is given as

f_smax = $\mu$_sN

f_smax = (0.56) (597.8)

f_smax = 334.8 N

f_s = actual static frictional force acting

F = applied force on the refrigerator = - 261 N                      (negative since it acts in negative x-direction)

force equation along the horizontal direction is given as

F + f_s = 0                                             (Since the refrigerator does not move)

- 261 + f_s = 0

f_s = 261 N

hence the static frictional force comes out to be 261 N in positive x-direction.

b)

the refrigerator starts moving when the applied force becomes equal to or greater than the maximum static frictional force.

hence F_applied = f_smax = 334.8 N