Question

A mother is pulling a sled at constant velocity by means of a rope at 37°. The tension on the rope is 120 N. Mass of children plus sled is 55 kg. The mother has a mass of 61 kg. Find the static friction acting on the mother.

Answer 1

f = 106.3 N

Step-by-step explanation:

The force applied on the sled must be equal to the static frictional force to move the sled:

Tension Force Horizontal Component = Static Frictional Force

`TCos\theta = \mu W\\TCos\theta = \mu mg`

where,

T = Tension = 120 N

θ = angle of rope = 37°

μ = coefficient of static friction = ?

m = mass of children plus sled = 55 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

`(120\ N)Cos\ 37^o = \mu (55\ kg)(9.81\ m/s^2)\\\\\mu = (95.84\ N)/((55\ kg)(9.81\ m/s^2))\\\\\mu = 0.18`

Now, the static friction acting on the mother will be:

`f = \mu mg = (0.18)(61\ kg)(9.81\ m/s^2)\\`

f = 106.3 N