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
[tex]\sum M = 0[/tex]
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.
[tex]\sum M_D_y = 0[/tex]
[tex]C_x = (120)-T(40)= 0[/tex]
sustitute 1433.33N for T
[tex]C_x (120)-(1433.33)(40)\\\\C_x (120)=57333.2\\\\C_x=477.78N[/tex]
The reaction at C along x-axis 477.78N
Apply the moment law of equilibrium at point D about x-axis
[tex]\sum M_D_z = 0[/tex]
[tex]-C_y (120)+860(80+120)=0\\\\-C_y(120)=-17200\\\\y=1433.33N[/tex]
The reaction at C along y-axis is 143.33N
Apply the force law of equilibrium along z direction.
[tex]\sum F_z=0[/tex]
[tex]C_z=0[/tex]
The reaction at C along z-axis is 0N
Apply the force law of equilibrium along x direction.
[tex]\sum F_x=0[/tex]
[tex]C_x+D_x+T=0[/tex]
substitute 477.78N for [tex]C_x[/tex] and 143.33N for [tex]D_x[/tex]
[tex]477.78N+D_x+1433.33N=0\\\\D_x=-1911.11N[/tex]
The reaction at D along x-axis is -1911.11 N
Apply the force law of equilibrium along y direction.
[tex]\sum F_y=0[/tex]
[tex]C_y+D_y-860=0[/tex]
substitute 1433.33 for [tex]C_y[/tex]
[tex]1433.33+D_y-860=0\\\\D_y=573.33[/tex]
