Answer:
a) the tension in the cord T is 1200 N
b)
reaction at C along x-axis Cy is
400 N
reaction at C along y-axis Cy is
1200 N
reaction at C along z-axis Cz is
0 N
reaction at D along X-axis Dx is
1600 N
reaction at D along y-axis Dy is
-480 N
Explanation:
a)
to find the tension in the cord, we say;
∑ Mz = 0
720(200) - T(120) = 0
144000 - 120T = 0
T = 144000/120
T = 1200 N
the tension in the cord T is 1200 N
b)
the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.
we make use of the moment law of equilibrium at point D about y-axis.
∑ MDy = 0
Cx(120) - T(40) = 0
we substitute value of T
Cx(120) - 1200(40) = 0
Cx 120 = 48000
Cx = 48000/120
Cx = 400 N
reaction at C along x-axis Cy is
400 N
Next we also apply the moment law of equilibrium at point D about x-axis.
∑MD z = 0
-Cy(120) + 720(80 + 120) = 0
-Cy(120) = - 144000
Cy = -144000 / - 120
Cy = 1200 N
reaction at C along y-axis Cy is
1200 N
we also apply the force law of equilibrium along z direction
∑Fz = 0
Cz = 0 N
reaction at C along z-axis Cz is
0 N
we also apply the force law of equilibrium along x direction
∑Fx = 0
Cx + Dx + T = 0
Substitute 400N for Cx and 1200 N for Dx
so
400 + Dx + 1200 = 0
Dx = 1600 N
therefore reaction at D along X-axis Dx is
1600 N
we also apply the force law of equilibrium along y direction
∑ Fy = 0
Cy + Dy - 720 = 0
we substitute Cy = 1200 N
so
1200 + Dy - 720 = 0
Dy = - 480 N
therefore reaction at D along y-axis Dy is
-480 N
