Guest

A block of mass 15 kgis placed on inclined plane when an angel of inclination is 30 the block just begin to slide the plane what is the sliding fricition

A block of mass 15 kgis placed on inclined plane when an angel of inclination is 30 the block just begin to slide the plane what is the sliding fricition

Grade:11

2 Answers

Arun
25750 Points
4 years ago
 
first of all 1 kg.wt means a force that is required to accelerate a mass of 15 kg with an acceleration of approx 9.8 m/s2.
here the block is just going to slide down the wedge..so friction force needed to stop the block is mg*sin30.
since, mass  = 15kg, force required  = 15*g*sin30 newton
= 7.5*g  newton
= 7.5 kg.wt …............... {since 1 kg.wt = 1*g newton, where g = 9.8 m/s2 }
Khimraj
3007 Points
4 years ago

In equilibrium,

F=mg sin 30°…………(1)

R=mg cos 30°…………(2)

On dividing ( 1) by (2), we get

F/R= mg sin 30°/mg cos 30°

or, friction coefficient= tan 30°.

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free