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 3030^{\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?

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After drawing a free body diagram, as shown below, with the pulling tension force of the rope TT acting at a θ=30\theta=30^{\circ} angle with the horizontal on the sled, the friction force FfF_f, weight FGF_G and normal reaction nn, we can project these forces on the vertical and horizontal axis and following Newton's first law have:

Tsinθ+nmg=0n=mgTsinθT\sin{\theta}+n-mg=0\Rightarrow n=mg-T\sin{\theta}

TcosθFf=0Ff=μn=μ(mgTsinθ)=TcosθT\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

μ=Tcosθ(mgTsinθ)=75cos(30)(609.875sin(30))=0.118\mu=\frac{T\cos{\theta}}{(mg-T\sin{\theta})}=\frac{75\cos(30)}{(60\cdot 9.8-75\sin(30))}=\boxed{0.118}

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