PHY 2053, Spring 2009, Quiz 5 — Whiting
1. a) What is the minimum force of friction required to hold the system shown in equilibrium?
Vertically: T − W1 = 0 ⇒ T = W1 if static.
Horizontally: T − FR = 0 ⇒ FR = T. So FR = W 1 = 50 N.
b) What coefficient of static friction between the 100-N
block and the table ensures equilibrium?
Vertically: N − W2 = 0 ⇒ N = W2 .
Now, FR = µs × N ⇒ µs = FR /N = W1 /W2 = 0.500.
c) If the coefficient of kinetic friction between the 100-N block and the table is 0.250, what
hanging weight should replace the 50.0-N weight to allow the system to move at constant
speed once it is set in motion?
As a = 0, we still have FR = W1 from a) but now FR = µk × N = µk × W2 from b). So
W1 = µk × W2 = 25 N.
2. A car is travelling at 50.0 km/hr on a flat highway.
a) If the coefficient of friction between the tires and the road on a rainy day is 0.100, what
is the minimum distance in which the car will stop?
Vertically: N − W = 0 ⇒ N = W. Horizontally: −FR = m × a, where FR = µk × N ⇒
a = −µk × g, since W = m × g. Then we can use 0 = v02 + 2a × ∆x to find:
v02
(50/3.6)2
∆x =
=
= 98.4 m.
2µk × g
2 × 0.100 × 9.8
b) What is the stopping distance when the surface is dry and the coefficient of friction is
0.600?
We can use the same equation to find:
v02
(50/3.6)2
∆x =
=
= 16.4 m.
2µk × g
2 × 0.600 × 9.8