Question

If a force of 86 lbs parallel to the surface of a 20 degree inclined plane will push a 120 lb block up the plane at a constant speed. What is the coeffficient of sliding friction

Answer 1

Question 32403

For an inclined plane,

1) $N = mg\cos \theta$

2) $ma = F - F_{f} - mg\sin \theta$

, where $F_{f}$ is the friction force, which is equal to $F_{f} = \mu N = \mu mg\cos \theta$ (according to first equation). $\mu$ Is the friction coefficient, one needs to find.

Hence, from equation 2):

$$a = \frac {F - \mu m g \cos \theta - m g \sin \theta}{m}.$$

Let us convert given force and mass into Newtons and kilograms:

$$m = 120 \, lb = 54.36 \, kg; \, F = 86 \, lbs = 382.2 \, N$$

Since the motion is with constant speed, acceleration is zero, which yields

$$F - \mu m g \cos \theta - m g \sin \theta = 0 \Rightarrow \mu = \frac {F - m g \sin \theta}{m g \cos \theta}, \text{ which after calculation gives } \mu \approx 0.399.$$