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In (Figure 1) a 6.00-m-long, uniform beam is hanging from a point 1.00 m to the right of its center. The beam weighs 155 N and makes an angle of 30.0∘ 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.If the system is in equilibrium, what is w ? You can ignore the thickness of the beam.

In Figure 1 a 600mlong uniform beam is hanging from a point 100 m to the right of its center The beam weighs 155 N and makes an angle of 300 with the vertical A class=

Respuesta :

Given::

Length = 6.00 m

Weight of beam = 155 N

Angle = 30.0 degrees

Weight added at the right-end = 100 N

If the system is in equilibrium, let's find the weight, w, at the left-end.

Since the system is in equilibrium, the net torque of the system will be zero.

Now, we have the equation:

[tex](100)(2.00sin30)-w(4.00sin30)-(155)(1.0sin30)=0[/tex]

Where w is the weight at the left end.

Let's solve for w.

We have:

[tex]\begin{gathered} (100)(1)-w(2)-(155)(0.5)=0 \\ \\ 100-2w-77.5=0 \\ \\ 100-77.5-2w=0 \end{gathered}[/tex]

Solving further:

[tex]\begin{gathered} 22.5-2w=0 \\ \\ 2w=22.5 \end{gathered}[/tex]

Divide both sides by 2:

[tex]\begin{gathered} \frac{2w}{2}=\frac{22.5}{2} \\ \\ w=11.25\text{ N} \end{gathered}[/tex]

Therefore, the weight hung at the left end is 11.25 N

ANSWER:

11.25 N

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