## Physics (10th Edition)

$\mu_s=0.665$
The image below shows the forces that act on the crate. We call the horizontal force that is mentioned in the exercise $P$ and analyze it into 2 components $P\cos\theta$ and $P\sin\theta$ First, let's consider the horizontal forces that affect the crate's motion: The crate is pulled down by its weight $mg\sin20$ and force $P\cos20$ and these are opposed by static friction $f_s$. Since $P=535N$ is required to start the crate, for $P=535N$, $f_s^{max}$ balances 2 opposing forces. $$mg\sin20+P\cos20=f_s^{max}=\mu_sF_N$$ $$\mu_sF_N=225\times9.8\sin20+535\cos20=1256.9N$$ To find $\mu_s$, we have to find $F_N$, which can be found by looking at the vertical forces. The vertical component of the crate's weight $mg\cos20$ is opposed by both $F_N$ and the vertical component of $P$, which is $P\sin20$. Since there is no vertical movement, these forces are balanced. $$mg\cos20=F_N+P\sin20$$ $$F_N=mg\cos20-P\sin20$$ $$F_N=225\times9.8\cos20-535\sin20=1889N$$ Therefore, $$\mu_s=\frac{1256.9N}{1889N}=0.665$$