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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.

Respuesta :

Answer:

f = 106.3 N

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

[tex]TCos\theta = \mu W\\TCos\theta = \mu mg[/tex]

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,

[tex](120\ N)Cos\ 37^o = \mu (55\ kg)(9.81\ m/s^2)\\\\\mu = \frac{95.84\ N}{(55\ kg)(9.81\ m/s^2)}\\\\\mu = 0.18[/tex]

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

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

f = 106.3 N