Question

A hiker with a mass of 85 kg wants to drag a 200 kg sled over snow. She does this by tying a rope to the sled (at ground level) and running it over her shoulder. The rope running between her and the sled makes a 45 degree angle with the horizontal. If the coefficient of friction between the sled and the snow is 0.1, what must the coefficient of friction between her boots and the ground be for her to move the sled?

Answer 1

Explanation:

According to the diagram, normal force acting on the hiker is as follows.

         $N_{1} = Mg + T Sin 45^{o}$

The frictional force acting on the hiker is as follows.

        $f_{1} = \mu \times N_{1}$

                  = $\mu (Mg + T Sin 45^{o})$

Normal force acting on the sled according to the figure is as follows.

         $N_{2} = mg - T Sin 45^{o}$

So, frictional force acting on the sled is as follows.

         $f_{2} = \mu' N_{2}$

      $T cos 45^{o} = \mu'(mg - T Sin 45^{o})$

               = $0.1 \times (85 kg \times 9.8 m/s^{2} - T Sin 45^{o})$

               = $83.3 - T Sin 45^{o}$

    $T cos 45^{o} + (0.1)T Sin 45^{o}$ = 83.3 N

              T = $\frac{83.3 N}{cos 45^{o} + (0.1)T sin 45^{o}}$

                 = $\frac{83.3 N}{0.707 + (0.1) \times 0.707}$

                 = 107.11 N

Frictional force acting on the sled is equal to the horizontal component of the tension in the rope and it is given as follows.

          $f_{1} = \mu (Mg + T sin 45^{o})$  

    $T cos 45^{o} = \mu (Mg + T sin 45^{o})$

          $\mu = \frac{T cos 45^{o}}{Mg + T sin 45^{o}}$

                    = $\frac{107.11 N \times 0.707}{85 \times 9.8 m/s^{2} + 107.11 \times 0.707}$

                   = $\frac{75.726}{833 + 75.726}$

                   = 0.08

Therefore, we can conclude that the coefficient of friction between her boots and the ground be for her to move the sled is 0.08.