Question

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 $30°$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?

Short answer

The co-efficient of friction is $\mu =0.118$

Step-by-step solution

Step 1. Given Information

We have,

walking at a steady $v=1.5$m/s

the rope pulls up on the sled at a $\theta =30°$angle

and with your friend on it, is $m=60$kg and that you're pulling a force of $T=75$N

Step 2. Find the friction

the free body diagram, as shown above, with the pulling tension force of the rope $T$acting at a $\theta =30°$angle with the horizontal on the sled, the friction $F_f$, weigh $F_G$and normal reaction$n$

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

we can express the coefficient of friction as

$$\begin{gathered} \mu =\frac{T\cos \theta }{(mg-T\sin \theta )} \\ \mu =\frac{75\cos \left(30\right)}{(60\times 9.8-75\sin (30\left)\right)} \\ \mu =0.118 \end{gathered}$$