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What force $\vec{F_{1}}$ is needed to balance the beam in the diagram below?

enter image description here

I know that $\sum \vec{F}$ must equal zero. I also know that since the unknown force is farther from the pivot, it will be of smaller magnitude than the 8.0N force acting on the opposite side. My question is, what calculations or formula can I use to solve this problem? A hint would be appreciated.

EDIT

Taking the adive from Spxn:

$\tau _{1}=\tau _{2}$

$\tau _{1}=F(l)$

$\tau _{1}=8(0.5)$

$\tau _{1}=4Nm$

$\tau _{2}=4Nm$

$4Nm = 0.75(F)$

$F = \frac{4Nm}{0.75}$

$F = 5.3N$

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  • $\begingroup$ A hint would be appreciated Here is the hint: In statics (when things are supposed to not move). There are essentially two rules that have to be fulfilled: $$\sum \vec F_{net} =0$$ and $$\sum \vec \tau_{net} =0$$. You are aware of $\sum \vec F_{net}$ I can see, so now look at $\sum \vec \tau_{net}$, where for each force produces $\tau = d F$. $\endgroup$
    – Steeven
    Jan 16, 2015 at 16:13

3 Answers 3

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your AIM : Balacing beam, so the object to be in the equilibrium condition. Conditions for equilibrium.

  • Net force = 0
  • Net Torque = 0

$$∑F = 0; \implies 8+F(cg)+F = 0;$$

Torque $$\sum \tau = 0;$$ calculating at CG.

$$8*0.5 - F*.75 = 0 \implies 4 - 0.75*F = 0; \implies 0.75*F = 4$$

$$F = \frac{4}{0.75} \implies F = 5.33~\text{N}$$

here comes net force equation again $$8+F(CG)+5.33 = 0 \implies F(CG) = -13.33~\text{N}$$ (negative sign indicates the direction of force at CG is in opposite direction )

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Hint: In this case, you would need the net torque to be zero.

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  • $\begingroup$ I agree aside from the point where you say $F_{net}$ does not have to be zero. You actually need $F_{net}$ to be zero, otherwise the beam will accelerate laterally. Don't forget the reaction force that the support exerts upwards! $\endgroup$
    – Involute
    Jan 16, 2015 at 15:03
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For these sorts of questions, you need to fulfil two conditions in order to satisfy equilibrium. If you don't satisfy equilibrium, things start to accelerate and move. The two conditions (i.e. equilibrium equations) are:

1) The net force, $F_{net}$ (a.k.a. resultant force) must be zero. This means the beam won't start accelerating away sideways, upwards, or downwards.

2) The net torque, $\tau_{net}$ must be zero. This means the beam does not start to rotate.

These two conditions alone will be enough to solve most problems involving an object that needs to be held stationary, provided the system is "statically determinate" (i.e. the equations of equilibrium are enough to solve the problem. If not, however, it is still possible to solve, in general, but relies on more complex concepts). The problem above is indeed statically determinate.

Condition 2) will be the best approach to answering your question here.

Also, be aware that the diagram can be quite misleading, because one of the forces is hidden: the support exerts an upward force on the beam. If the support was replaced with you holding the beam up, you would have to exert an upward force to keep the beam from falling down.

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