Question

A mule 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

Answer 1

(a) 1800 N

The equation of the forces along the vertical direction is:

$F sin \theta + N - mg = 0$

where

$F sin \theta$ 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

$\theta = 30.3^{\circ}$

We find N:

$N=mg-F sin \theta = (246)(9.8)-(1210)(sin 30.3^{\circ})=1800 N$

(b) 0.580

The equation of the forces along the horizontal direction is:

$F cos \theta - \mu_s N = 0$

where

$F cos \theta$ is the horizontal component of the push applied by the mule

$\mu_s N$ is the static frictional force

Substituting:

F = 1210 N

N = 1800 N

$\theta = 30.3^{\circ}$

We find $\mu_s$, the coefficient of static friction:

$\mu_s = \frac{F cos \theta}{N}=\frac{(1210)(cos 30.3^{\circ})}{1800}=0.580$

(c) 522 N

In this case, the force exerted by the mule is

$F= 6.05 \cdot 10^2 N = 605 N$

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

$F cos \theta - F_f = 0$

where

$\theta=30.3^{\circ}$

and $F_f$ 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 $\mu N$, depending on how much force is applied in the opposite direction)

Solving the equation,

$F_f = F cos \theta = (605)(cos 30.3^{\circ})=522 N$