200-mm lever and a 240-mm-diameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 720-N vertical load is applied at A when the lever is horizontal, determine (a) the tension in the cord, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.

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

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