Question

A 200-mm lever and a 240-mm-diameter pulley are welded to the axle BE that is supported by bearings at C and D. A 860-N vertical load is applied at A when the lever is horizontal. Assume that the bearing at D does not exert any axial thrust.

Answer 1

Answer:

A 200mm lever and a 240 mm diameter pulley are welded to the axle BE that is supported by bearings at C and D. If a 860N 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 axial thrust.

see explanation

Explanation:

Part a

The tension in the cord

$\sum M = 0$

860(200) - T(120) = 0

T(120) = 172000

T =1433.33N

The tension in the cord (T) = 1433.33N

Part b

Apply the moment law of equilibrium at point D about y-axis.

$\sum M_D_y = 0$

$C_x = (120)-T(40)= 0$

sustitute 1433.33N for T

$C_x (120)-(1433.33)(40)\\\\C_x (120)=57333.2\\\\C_x=477.78N$

The reaction at C along x-axis 477.78N

Apply the moment law of equilibrium at point D about x-axis

$\sum M_D_z = 0$

$-C_y (120)+860(80+120)=0\\\\-C_y(120)=-17200\\\\y=1433.33N$

The reaction at C along y-axis is 143.33N

Apply the force law of equilibrium along z direction.

$\sum F_z=0$

$C_z=0$

The reaction at C along z-axis is 0N

Apply the force law of equilibrium along x direction.

$\sum F_x=0$

$C_x+D_x+T=0$

substitute 477.78N  for $C_x$ and 143.33N for $D_x$

$477.78N+D_x+1433.33N=0\\\\D_x=-1911.11N$

The reaction at D along x-axis is -1911.11 N

Apply the force law of equilibrium along y direction.

$\sum F_y=0$

$C_y+D_y-860=0$

substitute 1433.33 for $C_y$

$1433.33+D_y-860=0\\\\D_y=573.33$