16. Two steel balls A and B are placed inside a right
circular cylinder of diameter 54
cm making
contact at points P, Q and R as shown. The radius
ra = 12 cm and rb
18 cm. The masses are ma
15 kg and 60 kg. The forces exerted by the floor at
the point Q and the wall at R are, respectively
(taking g= 10 m/s?)
54 cm
(a) 600 N, 150 N
(c) 600 N, 200 N
(b) 750 N, 150 N
d) 750 N, 200 N
Answers
☆ Answer ☆
- Two steel balls A and B are placed inside a right circular cylinder of diameter 54cm making contact at points P, Q and R as shown.
- The radius ra = 12 cm and rb 18 cm. The masses are ma 15 kg and 60 kg.
Have to Find :-
- The forces exerted by the floor at the point Q and the wall at R .
Solution :-
Now, as from the given, we have
- Circular cylinder of diameter 54.
- R(a) = 12 cm and R(b) =18 cm.
- Mass(a) = 15 kg and Mass(b) = 60 kg.
Therefore ,
Applying the formula,
F(g) =(mass of A ball + mass of B ball) ×g
Here,
F (g) = " Force Exerted "
M(a) = " Mass of the A ball "
M(b) = " Mass of the B ball "
g = " Gravitational Force "
Now, Substituting the values in the above formula, we get
= (60 + 15) × 10 = 750 N
Because the forces acting on the system of two balls are its weight and Normal reactions from floor and wall respectively.
Let's consider the force exerted by floor be " N " Newton.
Force according to the direction be
Force of upward = N
Force of downward = 750 N
Reason :- System of two balls is in equilibrium.
Therefore, Upward net force will become 0.
Then,
Force of downward -Force of upward = 0
Force of up = Force of down
F ( Down) = N = 750 N
So, we got the force exerted by the floor at the point of Q is 750 N.
━━━━━━━━━━━━━━━━━━━━━━━━
Now, Substituting the values in the above formula, we get
= (60 + 15) × 10 = 750 N
Because the forces acting on the system of two balls are its weight and Normal reactions from floor and wall respectively.
Let's consider the force exerted by floor be " N " Newton.
Force according to the direction be
Force of upward = N
Force of downward = 750 N Ans.