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Q. A body is placed on a rough inclined plane of inclination `theta`. As the angle...

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user8235304 | Student, Grade 11 | (Level 1) Valedictorian

Posted July 8, 2013 at 11:43 AM via web

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Q. A body is placed on a rough inclined plane of inclination `theta`. As the angle `theta` is increased from 0° to 90°; the contact force between block and plane:-

A)remains constant

B)first decreases then increases

C)first remains constant then decreases

D)first increases then decreases

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llltkl | College Teacher | (Level 3) Valedictorian

Posted July 8, 2013 at 12:59 PM (Answer #1)

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When a body is placed on a rough inclined plane of inclination `theta` , and the inclination is increased gradually from zero, the frictional force operates against the direction of motion. At first, when `theta` is zero, frictional force is zero. As `theta` increases slowly, the frictional force also increases proportionately until it reaches a certain limiting value called coefficient of static friction. When `theta` is quite small, `F_(nt)` cannot overcome static frictional resistance and the body remains stationary. As the angle `theta` is increased, at a point when `F_(nt)` surpasses the static frictional resistance, the body starts moving down the plane.

As soon as it starts moving, `mu_m` comes into play. The coefficient of friction (and hence frictional force) acting on it while it is in motion is `mu_m` which is less than `mu_s` , the coefficient of static friction. The net force in the direction of motion is given by:

`F_(nt)=mgsintheta-mu_m*mgcostheta`

As theta increases upto 90 degrees,` f=mu_m*mgcostheta` , decreases.

Therefore, the frictional contact force first increases, reaches a maximum, then decreases regularly with `theta` .

So, the correct answer is option D).

 

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