A ball and socket at the origin of coordinates and cables AB and AC support rigi

Question

A ball and socket at the origin of coordinates and cables AB and AC support rigid bar OA. Bar OA lies on the x-axis. The 495 N vertical force F loads the bar at its midpoint. (Due to the nature of this problem, do not use rounded intermediate values in your calculations-including answers submitted in WebAssign.) B(-4, 1, 6)m C(2, -4, 3)m9 BS A(4, 0, O)m - - =-(495 N) Determine the tension in each cable in newtons. TAB = Determine the reaction force in newtons at O. (Express your answer in vector form.)

Explanation / Answer

for the given Ball and socket at origin
let tension in each cable be Tac and Tab
considering i, j and k be unit vectors along x , y and z directions
in vector format
Tac = |Tac|((2-4)i + (-4-0)j + (3-0)k)/sqrt((2-4)^2 + (-4-0)^2 + (3-0)^2) = |Tac|(-2i - 4j + 3k)/sqrt(29)
similiarly
Tab = |Tab|(-8i + j + 6k)/sqrt(101)

let reaciton at socket joint = R i ( along +x axis)
then from force balance at point A

Tac + Tab + F + R = 0
|Tac|(-2i - 4j + 3k)/sqrt(29) + |Tab|(-8i + j + 6k)/sqrt(101) - 495k + Ri = 0
hence

-2Tac/sqrt(29) - Tab/sqrt(101) + R = 0
-4Tac/sqrt(29) + Tab/sqrt(101) = 0
Tab = 4Tac*sqrt(101)/sqrt(29)

3Tac/sqrt(29) + 6Tab/sqrt(101) = 495
Tac = 495*sqrt(29)/27 = 98.728 N

and
Tab = 4* 495*sqrt(101)/27 = 736.990878 N

R = 2Tac/sqrt(29) + Tab/sqrt(101) = 2*495/27 + 4* 495/27 = 110 N

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