Friction — Question 6

Question 6

Two surfaces make a moving sandwich

A 2 kg plate moves right at 2 m/s between two horizontal belts. The upper belt moves right at 5 m/s and exerts 8 N of kinetic friction on the plate. The lower belt is stationary and exerts 3 N of kinetic friction. Vertical forces balance. These magnitudes remain constant while both contacts slide.

1. Find the initial horizontal acceleration.

2. Find when the plate first matches one belt and how far it moves before that event.

3. What friction must the upper contact supply to maintain motion at 5 m/s afterward? State the required condition.

Question 6 - Solution

Two surfaces make a moving sandwich

(a) Assign each contact separately.

Evaluate slip at each contact: $$v_p-v_u=2-5=-3,\qquad v_p-v_l=2-0=+2\quad\mathrm{m/s}.$$ Hence the upper friction is $+8$ N and lower friction is $-3$ N: $$\boxed{a=\frac{8-3}{2}=2.5\ \mathrm{m/s^2}}.$$

(b) Find the first regime boundary.

The speed rises toward the upper belt speed. The first transition is $$2+2.5t_c=5\Longrightarrow\boxed{t_c=1.2\ \mathrm s}.$$ $$\boxed{\Delta x=2(1.2)+\tfrac12(2.5)(1.2)^2=4.2\ \mathrm m}.$$

(c) Replace the upper sliding law.

At a constant common speed of 5 m/s, lower contact still slides: $$f_{s,u}-3=ma=0\Longrightarrow\boxed{f_{s,u}=+3\ \mathrm N}.$$ Sticking to the upper belt is possible exactly when $f_{s,u,\max}\geq3$ N. Its old 8 N kinetic value does not determine this static capacity.