Question

In given figure all surfaces are smooth. The ratio of forces exerted by the wedge on mass M
when force 'F is not applied and when F is appled such that M is at rest with respect to wedge
[ ]
a) 1
b) 1:2
c) sec?e
d) cose​

Answer 1

Answer: Correct answer is option (c) $Sec^2$θ

Explanation:

Please refer to the attached diagram when going through the answer for better understanding.

When the ball moves, it's acceleration can be broken into two different components for calculations purpose,

We need to find the acceleration along the X-axis and Y-axis.

Acceleration along the x-axis will be

∑$F_x$ = 0 = R sinθ = ma

similarly acceleration along the y-axis will be

∑$F_y$ = 0 = R cosθ = mg

∴ R = $\frac{mg}{cos theta}$  = mg Secθ  .........................equation 1

After this the ball moves down the wedge, refer to the second figue in the attached image,

It's acceleration along x-axis will be - mg sinθ = ma

Ball moves along the slope and is moving in eqilibrium with $y^{'}$ aixs.

∴mg cosθ $R^'$ = 0

But we know, cos 0 = 1

∴ $R^'$  = mg cosθ   ............................ equation 2

We divide equation 1 by equation 2, we get

$\frac{R}{R^'}$ = $Sec^2$θ.

Hence, option (c) is the correct answer.

Answer 2

Given:

The mass of the wedge is M

The force acting on it is F

To find:

The ration of the force between the wedge and the ball.

Solution:

Let the mass of the ball be m

Now taking the horizontal and vertical component and the force acting on them.

F(x)=0=R.sinФ==m.a

F(y)=0=R.cosФ

=m.g

mg×secФ=R

Now using horizontal component F=0

mg.sinФ=ma

Vertical component F=0

mg.cosФ=R'

Now taking the ration of

R/R'=secФ/cosФ

R/R'=sec²Ф

So the correct option is c that is sec²Ф