Find the net torque and angular acceleration of the bar?

The drawing shows a bar lying on the surface of a horizontal table top. The bar is pivoted at the left end and is free to rotate with negligible friction in the horizontal plane of the table-top about the axis through the end. The bar is 2.0m long and has a mass of 0.60 kg.

The forces are as follows:
F1 = 14.0 N (at the pivot)
F2 = 12.0 N (at the 1.25m mark perpendicular to the bar)
F3 = 14.0 N (at the 2.0m mark)

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1 Answer
Feb 2, 2018

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Dividing forces F1F1 and F2F2 into perpendicular components.

Clearly,F1 cos 35F1cos35 and F3 cos 35F3cos35 can't produce rotation of the rod as,the angle between the force and distance vector is zero.

So,only sinsin components can cause rotation.

Now,considering torque about the pivoted point,and taking the direction of the torque coming out of the plane of the page to be positive, we can write,

-r_1*F_2 + r_2*F_3 sin 35 = I alphar1F2+r2F3sin35=Iα (where, II is the moment of inertia,for a rod about an axis passing through one end is (ml^2)/3ml23 and alphaα is the angular acceleration)

Given, r_1 =1.25,r_2=2,F_2=12,F_3=14r1=1.25,r2=2,F2=12,F3=14

So,net torque acting is (-15+16.06) N.m(15+16.06)N.m or, 1.06 N.m 1.06N.m Direction is along our chosen direction.

Now, 1.06 = (ml^2)/3*alpha1.06=ml23α

Given, m =0.60 ,l=2m=0.60,l=2

So, alpha = 1.325 (rad)/s^2α=1.325rads2

Note,here the pivoted point should have enough ability to pull the rod with a force of (F_1 cos 35+F_3 cos 35)(F1cos35+F3cos35) as there is no force to balance them.