Question

A patient lying at rest on an inclined bed starts to slide when the incline angle θ is to 17°.
What is the coefficient of static friction between the patient and the bed?
(Use g = 9.81 m/s2.)

Answer 1

Answer on Question #40923, Physics, Mechanics | Kinamatics | Dynamics

Question:

A patient lying at rest on an inclined bed starts to slide when the incline angle $\theta$ is to $17{}^{\circ}$ . What is the coefficient of static friction between the patient and the bed? (Use $g = 9.81 \, \text{m/s}^2$ .)

Answer:

Newton's laws of motion:

$x: \quad F_{fr} = mg \sin \theta$

$y: \quad N = mg \cos \theta$

Friction force equals $F_{fr} = \mu N = \mu mg\cos \theta$ , $\mu$ - coefficient of friction.

Therefore:

$$\mu = \frac {F _ {f r}}{m g \cos \theta} = \frac {m g \sin \theta}{m g \cos \theta} = \tan \theta \cong 0. 3 0 6$$

Answer: 0.306