Question

1)an object is placed on the surface of a smooth inclined plane of inclination θ.it takes time t to reach the bottom. if the same object is allowed to slide down inclined plane on inclination θ,it takes time nt to reach the bottom where n is a number greater than 1. what is coefficient?
2)The force required to just move a body up the inclined plane is doublethe force required to justprevent the body from sliding down the plane.the coefficient of friction is μ.find the inclination θ of the plane
3)a particle inside a hollow sphere of radius r, having coeffient of friction(1/√3), upto what height can particle be at rest?

Answer 1

Answer on Question #39297, Physics - Mechanics | Kinematics | Dynamics

Question:

1) An object is placed on the surface of a smooth inclined plane of inclination $\theta$. It takes time $t$ to reach the bottom. If the same object is allowed to slide down inclined plane on inclination $\theta$, it takes time $nt$ to reach the bottom where $n$ is a number greater than 1. What is coefficient?

2) The force required to just move a body up the inclined plane is double the force required to just prevent the body from sliding down the plane. The coefficient of friction is $\mu$. Find the inclination $\theta$ of the plane

3) A particle inside a hollow sphere of radius $r$, having coefficient of friction(1/V3), up to what height can particle be at rest?

Answer:

1) The first question is confusing, so we assume that objects are placed on the same height $h$ and the second included plane has the inclination $\theta_{1}$. Let us consider the first case. Objects has the acceleration $a = g \sin \theta$ and moves on the distance $\ell = h \tan \theta$. So,

$$\ell = \frac{a t^{2}}{2} = \frac{g \sin \theta t^{2}}{2} \Rightarrow t = \sqrt{\frac{2 h}{g \cos \theta}}.$$

In the same way one can obtain

$$n t = \sqrt{\frac{2 h}{g \cos \theta_{1}}}.$$

Whence

$$\frac{n t}{t} = \sqrt{\frac{2 h}{g \cos \theta_{1}}} \Bigg/ \sqrt{\frac{2 h}{g \cos \theta}} \Rightarrow n = \sqrt{\frac{\cos \theta}{\cos \theta_{1}}}.$$

2) The friction force is

$$F_{f} = \mu m g \cos \theta.$$

The force required to move a body up equals

$$F _ {1} = \mu m g \sin \theta + F _ {f}.$$

The force required to prevent a body from sliding down equals

$$F _ {2} = \mu m g \sin \theta - F _ {f}.$$

By conditions,

$$\frac {F _ {1}}{F _ {2}} = 2 = \frac {\mu m g \sin \theta + \mu m g \cos \theta}{\mu m g \sin \theta - \mu m g \cos \theta}$$ $$\sin \theta = 3 \cos \theta \Rightarrow \theta = \arctan 3.$$

3) The friction force equals

$$F _ {f} = \mu m g \cos \theta,$$

where $\mu = 1 / \sqrt{3}$, $\cos \theta = \frac{r - h}{r}$, $h$ is a height needs to be found. The friction force prevents a particle to slide down under the action of the component of the gravitational force

$$F _ {1} = m g \sin \theta.$$

So,

$$F _ {f} = F _ {1} \Rightarrow \mu m g \cos \theta = m g \sin \theta$$ $$\tan \theta = \mu \Rightarrow \theta = 30{}^{\circ}.$$ $$\cos \theta = \frac {\sqrt {3}}{2} = \frac {r - h}{r} \Rightarrow h = \frac {2 - \sqrt {3}}{2} r.$$