Please show legible work, will rate. A uniform horizontal strut weighs 340.0 N.

Question

Please show legible work, will rate.

A uniform horizontal strut weighs 340.0 N. One end of the strut is attached to a hinged support at the wall, and the other end of the strut is attached to a sign that weighs 1 80.0 N The strut is also supported by a cable attached between the end of the strut and the wall. Assume that the entire weight of the sign is attached at the very end of the strut. 30° Restaurant Fnd the tension in the cable in N) Find the force (in N) at the hinge of the strut. (Give the directien in degrees counterclockwise from the +x-axis. Assume that the x-axis is to the right.) magnitude direction counterclockwise from the +x-axis

Explanation / Answer

summing the moments about the hinge, and this one is no exception.

?M = 0 = T*sin30º*L - 340N*L/2 - 180N*L ? length L cancels
Tsin30 = 350 N
T = 700 N ? cable tension

At the hinge:
horizontally we have Tcos30 = Fx = 700N*cos30 = 606.217 N ? to the right on the end of the strut
vertically we have Fy = 350N + 180N - Tsin30 = 180 N ? up on the end of the strut

magnitude = ?(Fx^2+Fy^2 = ?(606.217^2+180^2) = 632.375N

direction = tan^-1 = 606.217/180 = 73.46deg

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