Equilibrium of a Rigid Body Class 11 Notes | EduRev

Physics Class 11

Class 11 : Equilibrium of a Rigid Body Class 11 Notes | EduRev

The document Equilibrium of a Rigid Body Class 11 Notes | EduRev is a part of the Class 11 Course Physics Class 11.
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6. Body is in equilibrium : -

We can say rigid body is in equillibrium when it is in

(a) Translational equilibrium

Equilibrium of a Rigid Body Class 11 Notes | EduRev

Fnet x = 0 and Fnet y = 0 and

(b) Rotational equilibrium

Equilibrium of a Rigid Body Class 11 Notes | EduRev 

Note :

(i) If net force on the body is zero then net torque of the forces may or may not be zero.

example. 

A pair of forces each of same magnitude and acting in opposite direction on the rod.

Equilibrium of a Rigid Body Class 11 Notes | EduRev

(2) If net force on the body is zero then torque of the forces about each and every point is same

t about B Equilibrium of a Rigid Body Class 11 Notes | EduRev

Equilibrium of a Rigid Body Class 11 Notes | EduRev

t about C Equilibrium of a Rigid Body Class 11 Notes | EduRev

Ex.18 Determine the point of application of third force for which body is in equillibrium when forces of 20 N & 30 N are acting on the rod as shown in figure 

Equilibrium of a Rigid Body Class 11 Notes | EduRev 

Sol. Let the magnitude of third force is F, is applied in upward direction then the body is in the equilibrium when

(i) Equilibrium of a Rigid Body Class 11 Notes | EduRev (Translational Equillibrium)

⇒ 20 + F = 30 ⇒ F = 10 N

So the body is in translational equilibrium when 10 N force act on it in upward direction.

(ii)

Let us assume that this 10 N force act. Then keep the body in rotational equilibrium So Tor que about C = 0

Equilibrium of a Rigid Body Class 11 Notes | EduRev

30 x 20 = 10 x 
x = 60 cm

Equilibrium of a Rigid Body Class 11 Notes | EduRev 

so 10 N force is applied at 70 cm from point A to keep the body in equilibrium.

Ex.19 Determine the point of application of force, when forces are acting on the rod as shown in figure.

Equilibrium of a Rigid Body Class 11 Notes | EduRev

Sol. Since the body is in equillibrium so we conclude Equilibrium of a Rigid Body Class 11 Notes | EduRev and torque about any point is zero i.e.,

Equilibrium of a Rigid Body Class 11 Notes | EduRev

Let us assume that we apply F force downward at A angle q from the horizontal, at x distance from B

Equilibrium of a Rigid Body Class 11 Notes | EduRev
⇒ Fnet x = 0 which gives
F2 = 8 N
From Fnet y = 0 ⇒ 5 + 6 = F+ 3
⇒ F1 = 8 N
If body is in equilibrium then torque about point B is zero,
⇒    3 x 5 + F1 x - 5 x 10 = 0
⇒    15 + 8x - 50 = 0
Equilibrium of a Rigid Body Class 11 Notes | EduRev 

Ex.20 A uniform rod length l, mass m is hung from two strings of equal length from a ceiling as shown in figure. Determine the tensions in the strings ? 

Equilibrium of a Rigid Body Class 11 Notes | EduRev 

Sol. Let us assume that tension in left and right string is TA and TB respectively. Then

Rod is in equilibrium then  Equilibrium of a Rigid Body Class 11 Notes | EduRev

From Equilibrium of a Rigid Body Class 11 Notes | EduRev
mg = TA + TB ...(1)
Equilibrium of a Rigid Body Class 11 Notes | EduRev
Equilibrium of a Rigid Body Class 11 Notes | EduRev

Equilibrium of a Rigid Body Class 11 Notes | EduRev 

Ladder Problems : 

Ex.21 A stationary uniform rod of mass `m', length `l' leans against a smooth vertical wall making an angle q with rough horizontal floor. Find the normal force & frictional force that is exerted by the floor on the rod? 

Equilibrium of a Rigid Body Class 11 Notes | EduRev 

Sol. As the rod is stationary so the linear acceleration and angular acceleration of rod is zero.

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