Question

A sphere of wight W = 100 N is kept stationary on rough inclined plane by a horizontal string AB as shown in figure . then :

• A. tension in the string is 100 N
• B. normal reaction on the sphere by the plane is 100 N
• C. tension in the string is $\frac{100}{2+\sqrt{3}}N$
• D. Force of friction on the sphere is $\frac{100}{2+\sqrt{3}}N$

Answer 1

Answer: (B), (C), (D)

The correct options are
B normal reaction on the sphere by the plane is 100 N
C tension in the string is $\frac{100}{2+\sqrt{3}}N$
D Force of friction on the sphere is $\frac{100}{2+\sqrt{3}}N$

It is given that inclined plane is rough, so there is friction.

Friction has an angle of ${30}^o$ with horizontal,

and it is perpendicular to normal, therefore angle between normal and vertical is ${30}^o$

see FBD of sphere.

since system is at rest so

Applying force balance in vertical direction:

$N\cos {30}^o+F\sin {30}^o=100$

$N\frac{\sqrt{3}}{2}+F\frac{1}{2}=100$

Applying force balance in horizontal direction:

$N\sin {30}^o=F\cos {30}^o+T$

$N\frac{1}{2}=F\frac{\sqrt{3}}{2}+T$

Applying torque balance at centre of sphere

$Fr-Tr=0$

$⟹F=T$

using this in above equation. $N\sin {30}^o=F\cos {30}^o+T$

$N\sin {30}^o=F\cos {30}^o+F$

$N\frac{1}{2}=F\frac{\sqrt{3}}{2}+F$

$⟹N=(2+\sqrt{3})F$

putting $N=(2+\sqrt{3})F$ in $N\frac{\sqrt{3}}{2}+F\frac{1}{2}=100$

$F(\sqrt{3}+2)=100$

$F=\frac{100}{\sqrt{3}+2}N$

since T=F prove above so

$T=\frac{100}{\sqrt{3}+2}N$

and $N=(2+\sqrt{3})F$

so $N=100N$