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I'm not sure how to calculate force $P$ when the moment is a given with the force $P$ at an angle.

I'm also having trouble breaking the moment into the $x$ and $y$ components.

force P causes a moment of 500 N-m about point A. Determine force P

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closed as off-topic by Algo, Chris Mueller, Wasabi, hazzey, Trevor Archibald Oct 26 '15 at 18:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about engineering, within the scope defined in the help center." – Algo, Chris Mueller, Wasabi, hazzey, Trevor Archibald
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ You need to resolve P into components normal to and parallel to the moment arm. $\endgroup$ – DLS3141 Oct 26 '15 at 16:53
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    $\begingroup$ Hi Neil, welcome to engineering.SE. While we don't forbid homework questions to be asked here, we do require that you make some effort to explain why you haven't been able to solve the problem on your own. What have you tried? Where are you stuck? $\endgroup$ – Chris Mueller Oct 26 '15 at 17:53
  • $\begingroup$ Hello Chris ,I couldn't get any examples of this type of problem. Everywhere I looked for examples it only showed how to determine moment with the forces already given, i tried to break the moment into a y- axis force and x-axis force and to devide the moment with the perpendicular distance from P. My textbook doesn't have the formula used below.That is what confused me. $\endgroup$ – Neil Venter Oct 26 '15 at 19:00
  • $\begingroup$ You have to go beyond "plug 'n chug". $\endgroup$ – DLS3141 Oct 27 '15 at 17:12
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$$ \tau = 500 \mbox{Nm} \\ $$

$$ \tau = r \times F \\ \tau = |r| |F| \sin{(\theta)} \\ $$

If the lower left angle is $43^{\circ}$, then the top angle from the bar $r$ to vertical is then $180 - 90 - 43 = \boxed{47^{\circ}}$. This means that $\theta$, the angle between the bar $r$ and the force $P$ is $47^{\circ} + 15^{\circ} = \boxed{\theta = 62^{\circ}}$.

Then,

$$ \tau = |r| |F| \sin{(\theta)} \\ |F| = \frac{\tau}{r \sin{(\theta)}} \\ |F| = \frac{ 500\mbox{Nm} }{(1.8 \mbox{m} ) \sin{(62^{\circ})}} \\ $$

$$ \boxed{F = 314.6 \mbox{N}} \\ $$

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