Question

Two steel balls a and b are placed inside a right circular cylinder of diameter 54cm making contact at point p q and r as shown in the figure the radius ra is equal to 12 cm and rb equal to 18cm. the massesa equals to 15 kg and Mb equals to 60 kg then find the force is exerted by the floor at the point of q (taking g=10m/s^2).

Answer 1

Answer:

750 N

Explanation:

1) The forces acting on the system of two balls are its weight and Normal reactions from floor and walls .

Now,

$F_g=(m_a+m_b)g=(60+15)*10=750 N$

And Force by floor is N.

Since,

System of two balls is in equilibrium.

Net upward force is 0.

F_up = N

F_down = 750.

Hence,

$F_{down}-F_{up}=0\\ \\ => F_{up}=F_{down}\\ \\=>N=750 N$

That is, Force exerted by the floor at the point of Q is 750 N.