Calculating Moments for F = 70 lb

[J-dizzal]

Homework Statement

For F = 70 lb, compute the combined moment of the two forces about (a) point O, (b) point C, (c) point D. The moments are positive if counterclockwise, negative if clockwise.

Homework Equations

M = F d , M = r x F

The Attempt at a Solution

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[J-dizzal]

Shouldt M = Fd here? where F=70 and d=37.014? i got an answer of 2590.95 but its not correct.

 

[haruspex]

Shouldt M = Fd here?

The image is a bit hard to read, but I assume the two forces are equal and opposite. M=Fd is right, where d is a certain distance. What distance? (It probably isn't the distance AB here.)

 

[gneill]

Shouldt M = Fd here? where F=70 and d=37.014? i got an answer of 2590.95 but its not correct.

Which part of the question are you trying to answer here? It looks like you're trying find the moment about point B, which isn't listed in the question statement. It asks for moments about points O, C, and D.

Also note that in order to use the form M = Fd you must be certain that the F is perpendicular to the d. Otherwise you should use the vector dot product or use geometry to find the component of F that is perpendicular to d.

 

[J-dizzal]

The image is a bit hard to read, but I assume the two forces are equal and opposite. M=Fd is right, where d is a certain distance. What distance? (It probably isn't the distance AB here.)

Yes are equal, opposite, parallel, and colinear.
Why is it not the distance AB, i thought the distance between couples was the distance for the formula M=Fd.

 

[J-dizzal]

Which part of the question are you trying to answer here? It looks like you're trying find the moment about point B, which isn't listed in the question statement. It asks for moments about points O, C, and D.

Also note that in order to use the form M = Fd you must be certain that the F is perpendicular to the d. Otherwise you should use the vector dot product or use geometry to find the component of F that is perpendicular to d.

if the two forces are parallel then the distance AB is perpendicular because point A and point B are each on the axis of either force

 

[haruspex]

colinear

Not collinear. If they were collinear there would be no moment.

i thought the distance between couples was the distance for the formula M=Fd.

It is, but the distance AB is the distance between two arbitrary points in the lines of application of the forces. It is not necessarily the distance between the lines.

 

[J-dizzal]

Not collinear. If they were collinear there would be no moment.

It is, but the distance AB is the distance between two arbitrary points in the lines of application of the forces. It is not necessarily the distance between the lines.

yes sorry, they are not on the same axis non collinear.

 

[J-dizzal]

Not collinear. If they were collinear there would be no moment.

It is, but the distance AB is the distance between two arbitrary points in the lines of application of the forces. It is not necessarily the distance between the lines.

i just tried using M=rxF and summed the two and got -2165.239$\hat k$
but that was wrong answer

 

[J-dizzal]

Which part of the question are you trying to answer here? It looks like you're trying find the moment about point B, which isn't listed in the question statement. It asks for moments about points O, C, and D.

Also note that in order to use the form M = Fd you must be certain that the F is perpendicular to the d. Otherwise you should use the vector dot product or use geometry to find the component of F that is perpendicular to d.

I thought the moment would be the same for any point? isn't that a properties of couple moments in 2d?

 

[J-dizzal]

Not collinear. If they were collinear there would be no moment.

It is, but the distance AB is the distance between two arbitrary points in the lines of application of the forces. It is not necessarily the distance between the lines.

well they are both points on the lines of application of the forces, i don't see how they are not the distance between the lines.

 

[haruspex]

well they are both points on the lines of application of the forces, i don't see how they are not the distance between the lines.

The distance between two parallel lines (or two skew lines ) is the shortest distance between points on them. If you pick two arbitrary points they are not necessarily as close as they could be, in fact they could be arbitrarily far apart.

 

[J-dizzal]

Not collinear. If they were collinear there would be no moment.

It is, but the distance AB is the distance between two arbitrary points in the lines of application of the forces. It is not necessarily the distance between the lines.

Ok i can see that i was assuming incorrectly that the line AB is prependi

The distance between two parallel lines (or two skew lines ) is the shortest distance between points on them. If you pick two arbitrary points they are not necessarily as close as they could be, in fact they could be arbitrarily far apart.

Ok, i was wrong to assume line AB was perpendicular to the couple. So would the best way to solve this would to be use ∑M = rxF from both the forces to point 0, then repeating that for the other two point?

 

[haruspex]

So would the best way to solve this would to be use ∑M = rxF from both the forces to point 0, then repeating that for the other two point?

Whichever way you do it will involve finding the perpendicular distance from a point to a line, so you might as well stick with your approach. You just need to do a bit of geometry to find that distance.

 

[J-dizzal]

Whichever way you do it will involve finding the perpendicular distance from a point to a line, so you might as well stick with your approach. You just need to do a bit of geometry to find that distance.

ok, but for future reference does the vector r in equation M= r x F have to be perpendicular to F?

 

[haruspex]

ok, but for future reference does the vector r in equation M= r x F have to be perpendicular to F?

No, the process of computing the cross product handles that. |r x F| = |r||F| sin(theta), where theta is the angle between the vectors.

 

[J-dizzal]

No, the process of computing the cross product handles that. |r x F| = |r||F| sin(theta), where theta is the angle between the vectors.

well why didnt my calculations work for M = r F

 

[haruspex]

well why didnt my calculations work for M = r F

It'll take me a while to tease through your numbers, but using AB as the distance vector and taking the cross product I get 2520. Most likely you have a sign error somewhere.

 

[J-dizzal]

It'll take me a while to tease through your numbers, but using AB as the distance vector and taking the cross product I get 2520. Most likely you have a sign error somewhere.

ok thanks that is correct, i just did it the long way by summing the two moments together using ∑M = rF = 2520.227 units

 

[J-dizzal]

It'll take me a while to tease through your numbers, but using AB as the distance vector and taking the cross product I get 2520. Most likely you have a sign error somewhere.

so the quicker way is to cross r_AB with F?

 

[J-dizzal]

so the quicker way is to cross r_AB with F?

yes, i confirmed that M = r x F . and it works for both vectors as long as r and F are head to head and not tail to head.

 

[J-dizzal]

 

[haruspex]

so the quicker way is to cross r_AB with F?

It's certainly quicker.

 

[J-dizzal]

It's certainly quicker.

indeed, i have an exam monday so i better remember this. thanks to all!