Question

One end of a uniform 4.30-m-long rod of weight F_g is supported by a cable at an angle of ? = 37

Answer 1

L = length of the rod = 4.3 m
Fg = weight of the rod = weight of the ball
? = angle of the cable = 37

Answer 2

along horzantal

N - T*cos? = 0

N = T*cos?

frictional force f = u*N = u*T*cos?

along vertical

Fnety = 0

f + T*sin? = Fg + Fg = 2Fg

u*T*cos? + T*sin? = 2Fg

u*T*cos? + T*sin? = 2Fg

T = Fg / (u*cos? + sin?)

net torque = 0

Fg*(L/2) + w*x = T*sin? *L

(L/2) + x = sin?*L / (u*cos? + sin?)

x = sin?*L / (u*cos? + sin?) - L/2

x = Fg*sin37*4.3 / (0.54*cos37 + sin37) - 2.15

2.15 + x = 2.505

x = 0.355m = 35.5 cm

Answer 3

along horzantal

N - T*cos? = 0

N = T*cos?

frictional force f = u*N = u*T*cos?

along vertical

Fnety = 0

f + T*sin? = Fg + Fg = 2Fg

u*T*cos? + T*sin? = 2Fg

u*T*cos? + T*sin? = 2Fg

T = Fg / (u*cos? + sin?)

net torque = 0

Fg*(L/2) + Fg*x = T*sin? *L = Fg*sin?*L / (u*cos? + sin?)

(L/2) + x = sin?*L / (u*cos? + sin?)

x = sin?*L / (u*cos? + sin?) - L/2

x = sin37*4.3 / (0.545*cos37 + sin37) - 2.15

2.15 + x = 2.505

x = 0.345m = 34.5 cm