Amule is harnessed to a sled having a mass of 246 kg, including supplies. the mule must exert a force exceeding 1210 n at an angle of 30.3° (above the horizontal) in order to get the sled moving. treat the sled as a point particle.
(a) calculate the normal force (in n) on the sled when the magnitude of the applied force is 1210 n. (enter the magnitude.) n
(b) find the coefficient of static friction between the sled and the fround beneath it.
(c) find the static friction force (in n) when the mule is exerting a force of 6.05 ✕ 102 n on the sled at the same angle. (enter the magnitude.) n

(a) 1800 N

The equation of the forces along the vertical direction is:

where

is the component of the applied force along the vertical direction

N is the normal force on the sled

mg is the weight of the sled

Substituting:

F = 1210 N

m = 246 kg

We find N:

(b) 0.580

The equation of the forces along the horizontal direction is:

where

is the horizontal component of the push applied by the mule

is the static frictional force

Substituting:

F = 1210 N

N = 1800 N

We find , the coefficient of static friction:

(c) 522 N

In this case, the force exerted by the mule is

So now the equation of the forces along the horizontal direction can be written as

where

and is the new frictional force, which is different from part (b) (because the value of the force of friction ranges from zero to the maximum value , depending on how much force is applied in the opposite direction)

Solving the equation,

v=ir = 3x15 =45 v