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The question I'm working on is:

A non-uniform beam AB of mass 5kg rests horizontally in equilibrium, supported by two light strings attached to the ends of the beam. The tensions in the strings are T1 and T2 and the strings make angles of 30" and 40" with AB and BA respectively. Find the magnitudes of T1 and T2.

I just want to make sure that I have solved vertically, horizontally and the moments correctly, since I have been struggling on this one for a while now.

Setting the bar length to $l$ and the distance to the center of gravity to $x$, so far I have...

Vertically:

$$ T_1 sin \theta_1 + T_2 sin \theta_2 - mg = 0 $$

Horizontally:

$$ T_1 cos \theta_1 - T_2 cos \theta_2 = 0 $$

Moments about A:

$$ T_2 l sin \theta_2 - xmg = 0 $$

Moments about B:

$$ T_1 l sin \theta_1 - (l-x)mg = 0 $$

But I can't seem to find a value for $T_1$ or $T_2$, unless I have it as a function for $x$. Have I made a mistake with my equations, or have I just missed something in the solution?

Thanks

Jbo
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    You shouldn't actually need the moment equations here, look again at your two force equations. This gives you a $2 \times 2$ system of linear equations in terms of $T_1$ and $T_2.$ – Stephen Donovan Apr 21 '21 at 17:16

1 Answers1

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Your first equation says, "the sums of the vertical components of the the forces on the beam equal zero", which is correct because the beam is in equilibrium.

The second equation is similar, with "vertical" replaced by "horizontal".

But there is no need to introduce variables $\theta_1$ and $\theta_2$, since we know that $\theta_1 = 30^\circ$ and $\theta_2 = 40^\circ.$ So really, the equations are:

Vertically:

$$ T_1 sin 30^\circ + T_2 sin 40^\circ - mg = 0 $$

Horizontally:

$$ T_1 cos 30^\circ - T_2 cos 40^\circ = 0 $$

and now this is just two equations with two unknowns, $T_1$ and $T_2.$

Adam Rubinson
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