Express the moment (torque) as a function of theta

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SUMMARY

The discussion centers on calculating the torque (moment) exerted by a force of 150 lb applied to a wrench at a distance of 43 inches horizontally and 6 inches vertically from point P. The user initially calculated moments for specific angles (0°, 30°, 60°, and 90°) using the cross product of the position vector and the force vector but received incorrect results. After clarification on the force's vertical application and the need to graph the moment as a function of theta, the user successfully submitted values for every 10 degrees of theta, discovering a non-linear relationship between moment and theta.

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tentoes
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Homework Statement



The problem shows a wrench with a force being applied to one end. Rotation will occur around point P at the opposite end. The force is applied 43 inches to the left from P, and 6 inches above P. The angle theta is between the vertical axis at the end of the wrench (non-p end) and the force being applied - which = 150lb. The problem requires solving for the moment in terms of theta and graphing Mp vs theta. I feel like this should be a super easy problem but I've already given three incorrect answers.

I took the cross product of vector r =<-3.58, 0.5> ft and F = <150cos(theta), 150sin(theta)> and then plugged in values of theta to the result to get that theta = 0, Mp = 537lb/ft, theta = 30, Mp = 502.5, theta = 60, Mp = 333, theta =90, Mp 75. I have no idea what I'm doing wrong here. I also just tried calculating the components of the moment vector and then getting the magnitude for different values of theta by doing sum of the squares...what am I missing?

Just to clarify, the angle theta given in the picture requires that the x component of the force vector use sin(theta), and the ycomponent use cos(theta). Clockwise is supposed to be the positive direction. I feel like this isn't even a question, I have no idea how I could be getting this incorrect.
 
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Not quite clear. Is the non-P end of the wrench the point where the force is applied? Or is the wrench horizontal?
 
Yeah - the wrench is horizontal, the P end of the wrench where the moment is applied is on the right, the force is applied in an upward direction on the left end of the wrench. Clockwise is positive for this problem, so I think no matter what the angle theta, a positive / clockwise moment will exist. I also tried the above answer backwards (for some reason) and that was wrong too.
 
tentoes said:
Yeah - the wrench is horizontal, the P end of the wrench where the moment is applied is on the right, the force is applied in an upward direction on the left end of the wrench. Clockwise is positive for this problem, so I think no matter what the angle theta, a positive / clockwise moment will exist. I also tried the above answer backwards (for some reason) and that was wrong too.
Then I don't understand the "six inches above P" part. How can the force be applied at a point not on the wrench? Please try posting a diagram, or describe the arrangement very clearly.
 
Sorry - usually I can't copy information for the problems but this one allows it -
Probs.4-19_20.jpg
 
Here's the whole problem statement: "Determine the torque (moment) MP
render?expr=M_P.gif
that the applied force F
render?expr=F.gif
= 150 lb
render?expr=%7B%5Crm+lb%7D.gif
exerts on the pipe about point P
render?expr=P.gif
as a function of θ
render?expr=%5Ctheta.gif
. Plot this moment MP
render?expr=M_P.gif
versus θ
render?expr=%5Ctheta.gif
for 0∘≤θ≤90∘
render?expr=0%5E%5Ccirc+%5Cle+%5Ctheta+%5Cle+90+%5E%5Ccirc.gif
. Consider positive moment as clockwise."
 
Ok, the picture helped. Your values for theta = 0 and 90 are easily seen to be correct, and the others look reasonable. You say you have to graph it, but (some automated checker?) tells you your answer is wrong. You can't feed it the whole graph, so I'm guessing you're asked to enter the moments for those four particular values of theta, right?
 
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It has a graphing app but the values listed on the x-axis are theta = 0, 30, 60, 90 so those are the only ones I entered values for. There are tick marks for 10 degree increments of theta - if you think I did it right maybe I could try adding the values at those tick marks and see if it accepts that - it's really weird.
 
OK - I just submitted it using values for every 10 degree of theta and it accepted it - I think the point of that was that the value for the moment at P is actually higher when theta = 10 than theta = 0, and the relationship between moment and theta is NOT linear - which it sort of was when I just plugged in the four values. Whew - thanks for your time!
 
  • #10
tentoes said:
OK - I just submitted it using values for every 10 degree of theta and it accepted it - I think the point of that was that the value for the moment at P is actually higher when theta = 10 than theta = 0, and the relationship between moment and theta is NOT linear - which it sort of was when I just plugged in the four values. Whew - thanks for your time!
Glad you found the magic trick.
 

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