Question

two men support A uniform horizontal beam of mass M at its two end if one of them suddenly let's go the force exerted by the beam on the other men will be

a)mg/2
b)mg/3
c)mg/4
d)mg/5
​

Answer 1

• The force exerted by the beam on the other men will be mg/4.

Given-

• Mass of beam = M

Initial and final condition of the beam is shown in the attached figure-

From the initial figure it is very clear that-

2 R = W so

R = W/2

In the final situation when one men suddenly go off then two types of motion takes place-

(i) Translational motion

(ii) Rotational motion

For translational motion -

F net = ma

W - R' = W/g × a        --------------------(eq. (i) )

R' is the new force that is exerted by the beam.

For rotational motion-

Torque net = I α , where I is the moment of inertia.

Moment of inertia for this beam = ML²/ 3

There will be no torque by men.

Torque will be only for the weight of the beam.

τ = F × R

τ = F R sin Ф

here Ф = 90° so sin Ф = 1

τ = F R

τ = MgL/2 = I α

MgL/2 = ML²/3 α

α = 3g/2L

For rotational motion-

a = R α

a = L/2 α

a = L/2 × 3g / 2L

a = 3 g /4

By putting the value of a in equation (i) we get -

W- R' = W/g × 3g / 4

R' = W - 3W/4

R' = W/4 = Mg/4

Hence R' < R

Regards