It’s a snowy day and you’re pulling a friend along a level road on a sled. You’ve both been taking physics, so she asks what you think the coefficient of friction between the sled and the snow is. You’ve been walking at a steady 1.5 m / s, and the rope pulls up on the sled at a $30^{\circ}$ angle. You estimate that the mass of the sled, with your friend on it, is 60 kg and that you’re pulling with a force of 75 N. What answer will you give?

After drawing a free body diagram, as shown below, with the pulling tension force of the rope $T$ acting at a $\theta=30^{\circ}$ angle with the horizontal on the sled, the friction force $F_f$, weight $F_G$ and normal reaction $n$, we can project these forces on the vertical and horizontal axis and following Newton's first law have:

$$T\sin{\theta}+n-mg=0\Rightarrow n=mg-T\sin{\theta}$$

$$T\cos{\theta}-F_f=0\Rightarrow F_f=\mu n=\mu (mg-T\sin{\theta})=T\cos{\theta}$$

We can express the coefficient of friction as

$$\mu=\frac{T\cos{\theta}}{(mg-T\sin{\theta})}=\frac{75\cos(30)}{(60\cdot 9.8-75\sin(30))}=\boxed{0.118}$$

$$\sum{F}_y = 0$$

$$T \sin( 30 \text{\textdegree} ) + n - mg =0$$

$$n = mg - T \sin( 30 \text{\textdegree} )$$

$$n = \left( 60\; \mathrm{kg} \right) \left( 9.8 \;\dfrac{ \mathrm{m} }{ \mathrm{s^2} } \right) - \left(75 \; \mathrm{N} \right) \left( \sin(30 \text{\textdegree}) \right)$$

$$\boldsymbol{ n = 551.1 \; \mathrm{N} }$$

$$\sum{F}_x =0$$

$$T \cos(30 \text{\textdegree} ) - f_k = 0$$

$$T \cos(30 \text{\textdegree} ) = f_k$$

$$T \cos(30 \text{\textdegree} ) = \mu n$$

$$\mu = \dfrac{ T \cos(30 \text{\textdegree} ) }{ n }$$

$$\mu = \dfrac{ \left(75 \; \mathrm{N} \right) \left( \cos(30 \text{\textdegree} ) \right) }{ 551.1 \; \mathrm{N} }$$

$$\boldsymbol{ \boxed{ \mu = 0.118 } }$$