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Assuming the bar Is uniform in density.

My idea was: since the system is in both translational and rotational equilibrium.

$T\sin\theta = F$ (Pivot on Bar)

Solving for $T$

let M = mass of bar

let m = mass of block

$T\cos\theta*l = Mg*\frac{1}{2} + mg$

$T = \frac{Mg\frac{1}{2} + mg}{\cos\theta*l}$

Then

$F$ (Pivot on Bar) = $T\sin\theta$

Although when I attempt the examples I don't receive the right answer.

thanks for any help out there.

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  • $\begingroup$ Which member, horizontal or vertical, member is the “bar” $\endgroup$
    – Bob D
    Commented Oct 2, 2019 at 3:38
  • $\begingroup$ @BobD the horizontal bar holding the block mass $\endgroup$
    – Bdyce
    Commented Oct 2, 2019 at 3:39
  • $\begingroup$ Ok. Have you taken a class in statics? This is a statics problem $\endgroup$
    – Bob D
    Commented Oct 2, 2019 at 4:34
  • $\begingroup$ @BobD I am taking that class right now, I just can't seem to find the solution for this. $\endgroup$
    – Bdyce
    Commented Oct 2, 2019 at 4:35
  • $\begingroup$ OK, see my answer $\endgroup$
    – Bob D
    Commented Oct 2, 2019 at 4:45

1 Answer 1

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We don't solve homework and exercise problems on this site. So I will only give you guidance.

You are going about it in the wrong way. You can't determine the tension in the string the way you are doing. It assumes the string supports the entire load, when it doesn't. There is also the horizontal and vertical components of the reaction of the vertical support at the pin.

Do a free body diagram on the bar. The pivot has no moment resistance (resistance to torque). Take the sum of the moments about the pivot and set equal to zero. That will let you calculate the tension in the string.

Then take the sum of the horizontal and vertical forces on the bar, which should include the contributions of the horizontal and vertical component of the tension force as well as the horizontal and vertical reactions at the pivot and as well as the weights and set them to zero. You can now calculate the reactions at the pivot.

Good luck and hope this helps.

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  • $\begingroup$ I am confused as the only torque forces exerted on the bar are the y component of the string, the bar mass, and the block mass if we take the axis of rotation to be at the pivot. Is this not correct? That is the equation I am solving for in my question. Sorry about my confusion It's been a long one today. Thank you for the help. $\endgroup$
    – Bdyce
    Commented Oct 2, 2019 at 4:54
  • $\begingroup$ I didn’t say they were the ONLy ones. I wanted to make sure you included them along with the contributions of the weights. Sorry if you misunderstood. I will clarify $\endgroup$
    – Bob D
    Commented Oct 2, 2019 at 5:52

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