# Engineering

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Calculate the moment about a point (5; 4; 3) ft. caused by the forces
P = 3^i - 2^j - 4^k lb. and Q = 5^i + 7^j - 2^k lb. acting at the points
(7; 2; -2) ft. and (-3; -2; 5) ft., respectively.

vector r1= (7,2,2) - (5,4,3) = 2i-2j-5k

vector r2= (3,2,5) - (5,4,3) = -8i-6j+2k

Vector M1= (2i-2j-5k) X P = -7i-7j+2k
Magnitude= 7.55 ft-lbs

Vector M2= (-8i-6j+2k) X Q = -2i-6j+85k
Magnitude= 86.23 ft-lbs

Moment about the given point = (7.55) + (86.23) = 93.78 ft-lbs

I'm unsure if this is right
No one has answered this question yet.

• Engineering -

You seem to have dropped some minus signs, but then corrected the errors.

vector r1= (7,2,-2) - (5,4,3) = 2i-2j-5k

vector r2= (-3,-2,5) - (5,4,3) = -8i-6j+2k

I do not agree with your cross products.
M1 should be
| i j k|
|2 -2 -5|
|3 -2 -4|
The i component is 8 -10 = -2
The j component is -15 +8 = -7
The k component is -4 +6 = 2

Are you sure you didn't mean M1 = -2i -7j +2k ?

The magnitude of -7i-7j+2k would not be 7.55.

So far, You seem to be going about it the right way but making typo errors.

For the total moment about the point, subtract the components first. Then take the sqrt of the sum of the squares of the differences. That is not what you did. You added magnitudes of vectors that are not parallel.

Thanks you for showing your work. I wish more students would do that.

• Engineering -

Thank you, I reevaluated my work and found all the errors.

Though I'm not sure sure why you would subtract M1 and M2 if you are looking for the total force applied?

Wouldn't M1 + M2 make more sense in this case?

• Engineering -

Such as (2i-7k+2k) + (-2i-6j+26k)?

To get to get my final vector in which I take the squareroot of the sum of the squares?

Also thanks I didn't recognize my conceptual error.

But I still believe it would be M1 + M2.

• Engineering -

You cannot algebraically add magnitudes unless they are in the same direction. M1 and M2 are not.

Add the two moments as vectors and THEN compute the magnitude.

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