Friction — Practice Problem Solutions
1. Kinetic friction: $F_{f,k}=\mu_kF_N$ — this gives the exact friction force while the object is already sliding. Static friction: $F_{f,s}\le\mu_sF_N$ — this gives the maximum possible value; the actual static friction adjusts to match whatever is needed to keep the object still, up to that ceiling.
2. $F_N=mg=(8.0)(9.8)=78.4\text{ N}$. $F_{f,k}=\mu_kF_N=(0.30)(78.4)\approx23.5\text{ N}$.
3. The crate doesn't move, so the actual static friction exactly balances the applied force: $40\text{ N}$. Maximum possible static friction $=\mu_sF_N=(0.50)(12)(9.8)=58.8\text{ N}$. Since the actual friction (40 N) is well below the maximum (58.8 N), the crate is not close to slipping — there's nearly 19 N of margin left.
4. They experience the same kinetic friction either way. Friction depends only on $\mu_k$ and the normal force (equal to the block's weight in both orientations) — not on how much surface area is in contact.
5. $F_{f,k}=\mu_kmg=(0.15)(20)(9.8)=29.4\text{ N}$.
6. a) Static friction increases to exactly match the growing applied force (keeping the book motionless) right up until it reaches its maximum value, $\mu_sF_N$ — at that point it can't increase any further, and the book begins to slide. b) Because $\mu_s$ is generally larger than $\mu_k$, the friction force often drops slightly the instant sliding begins — kinetic friction takes over, and its coefficient is typically a bit lower than the maximum static value that was resisting the book just before it slipped.
7. a) The A-B interface friction supports only Box A's weight: $f_1=\mu_1mg$. b) The B-floor interface friction supports the weight of both boxes: $f_2=\mu_2(m+3m)g=4\mu_2mg$. c) Adding a third block of mass $m$ on top of Box A increases the normal force at both interfaces. At A-B, the normal force becomes $(m+m)g=2mg$, so $f_1$ doubles to $2\mu_1mg$. At B-floor, the normal force becomes the total weight of all three blocks plus Box B, $(m+m+3m)g=5mg$, so $f_2$ increases to $5\mu_2mg$. Both friction forces increase because both interfaces now support more weight.
8. (B) — friction increases to match the applied force until it reaches $\mu_sF_N$, then drops to the lower, constant value $\mu_kF_N$ once sliding begins.