A force F = 7^i + 6^j - 5^k lb. acts at the origin. Find the magnitude and
direction cosines of the moment of this force about point A (-4; 5; 6) ft
So in this case wouldn't
vector r= -4i +5j +6k
and
vector F = 7i + 6j - 5k lb?
and r X F = -61i +22j -59k = MA
[MA]= 87.67 ft-lbs
cosα = -61/87.67 => α=134°
cosß= 22/87.67 => ß=75.5°
cosγ= -59/87.67 => γ=132.3°
Not sure exactly where I went wrong because those angles don't make any sense.
If the force acts at the origin, the lever arm r about point A is the opposite vector,
r = 4i -5j -6k
Ah, I knew there must be something off about my answer.
Thank you so much!!!
To find the moment of force about a point A, we need to calculate the cross product of the position vector r from point A to the origin with the force vector F.
In this case, the position vector r = -4i + 5j + 6k ft and the force vector F = 7i + 6j - 5k lb.
To calculate the cross product r x F, we can use the formula:
r x F = (r2*F3 - r3*F2)i - (r1*F3 - r3*F1)j + (r1*F2 - r2*F1)k
Substituting the values, we get:
r x F = (-5*(6) - 6*(-5))i - (-4*(-5) - 6*7)j + (-4*6 - 5*(-4))k
= (-61)i + (22)j - (59)k
So the moment of force about point A (MA) is -61i + 22j - 59k ft-lb.
Now, to find the magnitude of MA, we can use the magnitude formula:
|MA| = sqrt((-61)^2 + (22)^2 + (-59)^2)
= sqrt(3721 + 484 + 3481)
= sqrt(7690)
≈ 87.67 ft-lb
So the magnitude of the moment of force about point A is approximately 87.67 ft-lb.
To find the direction cosines of the moment of force, we can divide each component of the moment vector by its magnitude:
cosα = -61 / 87.67 ≈ -0.697
cosβ = 22 / 87.67 ≈ 0.251
cosγ = -59 / 87.67 ≈ -0.672
Now, to find the angles α, β, and γ, we can take the inverse cosine (or arccosine) of the direction cosines:
α ≈ arccos(-0.697) ≈ 134°
β ≈ arccos(0.251) ≈ 75.5°
γ ≈ arccos(-0.672) ≈ 132.3°
So the direction cosines of the moment of force about point A are approximately cosα = -0.697, cosβ = 0.251, and cosγ = -0.672, which correspond to angles α ≈ 134°, β ≈ 75.5°, and γ ≈ 132.3°, respectively.