Force Vectors
If the resultant force acting on the eyebolt is 1600 N and its direction measured clockwise from the positive x-axis is 30°, determine force (F1) & angle (φ) for the shown figure.
SOLUTION :
Let us use u for “phi” symbol.
Now,
∑Fx = 0
=> F1 cos(u) + 500 cos(60) - 450 * 3/5 - 1600 cos(30) = 0
=> F1 cos(u) - 1405.64 = 0
=> F1 cos(u) = 1405.64 (N) .......... (1)
∑Fy = 0
=> 1600 sin(30) + F1sin(u) - 500 sin(60) - 450 * 4/5 = 0
=> F1 sin(u) + 6.99 = 0
=> F1 sin(u) = - 6.99 …………… (2)
Doing (1) / (2), we get, cot (u) = 1405.64/(-6.99) = - 201.70
=> u = phi = acot (- 201.70) = - 0.2848º
Direction of F1 is at 0.2848º measured clockwise from x-axis
(ANSWER).
Now,
F1 = 1405.64 / cos(-0.2848) = 1405.66 (N) (ANSWER)
SOLUTION :
Let us use u for “phi” symbol.
Now,
∑Fx = 0
=> F1 cos(u) + 500 cos(60) - 450 * 3/5 - 1600 cos(30) = 0
=> F1 cos(u) - 1405.64 = 0
=> F1 cos(u) = 1405.64 (N) .......... (1)
∑Fy = 0
=> 1600 sin(30) + F1sin(u) - 500 sin(60) - 450 * 4/5 = 0
=> F1 sin(u) + 6.99 = 0
=> F1 sin(u) = - 6.99 …………… (2)
Doing (1) / (2), we get, cot (u) = 1405.64/(-6.99) = - 201.70
=> u = phi = acot (- 201.70) = - 0.2848º
Direction of F1 is at 0.2848º measured clockwise from x-axis
(ANSWER).
Now,
F1 = 1405.64 / cos(-0.2848) = 1405.66 (N) (ANSWER)
If the resultant force acting on the eyebolt is 1600 N and its direction measured clockwise from the positive x-axis is 30°
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