91影视

Given that a 6.00-m-long, uniform beam is hanging from a point 1.00 m to the right of its center. The beam weighs 140 N and makes an angle of with the vertical. At the right-hand end of the beam a 100.0-N weight is hung; an unknown weight w hangs at the left end.

Weight of beam, $w_1=140\mathbf{N}$

Weight hung on the right end $w_2=100\mathbf{N}$

Moment of arm for role="math" localid="1668071757315" $w_1\mathrm{is}1\mathbf{m}$

Moment of arm for${\mathrm{w}}_2\mathrm{is}2\mathbf{m}$

Torque $蟿=FI$

Where Fis force exerted and l is moment arm.

Let w be unknown weight, the moment of arm for w is 4m

Apply first condition for equilibrium.

Net torque on the beam is zero.

This implies

$w\left(4\mathbf{m}\right)\sin 鈦�{30}^{鈭�}+\left(140\mathbf{N}\right)\left(1\mathbf{m}\right)\sin 鈦�{30}^{鈭�}=\left(100\mathbf{N}\right)\left(2\mathrm{m}\right)\sin 鈦�{30}^{鈭�}$

The common factor$\sin 30掳$ cancels out,

w = 15 N

Thus weight is 15 N.

When the angle is increased to $45掳$instead of $30掳$then the common factor of cancels out and weight remains the same.

Hence w = 15N