An object of mass 20kg slides down an inclined plane at an angle of 30^o to the horizontal. The coefficient of static friction is?
[g = 10ms^-2]

FlynnRider

5 years ago

no it isn't.

let's follow what you said..

mg sin0/mgcos0


20×10×sin30/20×10×cos30.


=100/173


= 0.57

approximately 0.6 bro.

kennyddon

10 years ago

static friction is mgsin0/mg while coeficient of frictional force is mgsin0/mgcos0

Ogbonna_Chisom

3 years ago

Given:

Mass m=20 kg

Angle θ=30∘


The static friction is the frictional force that opposes an object to move. This means that in an incline, the frictional force must be equal to the force of gravity. In equation form

Fg=f

where Fg

is the gravitational force and f is the static frictional force.

In an incline plane, those two forces are given by

mgsinθ=μsmgcosθ

Since both the mass and acceleration due to gravity exist on both sides of the equation, it cancels out. Therefore

μs=sinθ/cosθ

Substitute the given values, we obtain

μs=sin(30)/cos(30)

μs=0.58

jesusthehealer

10 years ago

Mu=frictionalforce/normalreaction(R)=mg

agadido

3 years ago

na all of u dey fall for this.
na tan theta that's all

Ellaoba123

2 months ago

how do we do tan @ in jamb na 😭

kennyddon

10 years ago

d ansa is 0.5

Amverified

5 years ago

C=F/U
C=?
F=mgsin30°:20x10sin30°
F=100N
U=mg:20x10
U=200N
C=100/200
C=0.5
And is C