Question

One end of a uniform rod of weight w = 72.5 N and length L = 2.80 m is supported by a cable at an angle of theta = 33.5 degree above the rod. The other end rests on a small frictionless support and presses into a wall as shown in the figure. Determine the magnitude n of the vertical normal force exerted by the support on the rod and find the magnitude f of the tension in the cable. n = N T = N

Answer 1

うし

Weight of the rod = W = 72.5 N

Length of the rod = L = 2.8 m

Tension in the cable = T

Angle the cable makes with the horizontal = $\theta$ = 38.5^o

Vertical normal force exerted by the support = n

The rod is uniform therefore the center of mass of the rod will lie at the center of its length.

Taking moment balance about the left end of the rod,

W(L/2) = LTSin$\theta$

(72.5)(1.4) = (2.8)TSin(38.5)

T = 58.23 N

Balancing all forces in the vertical direction on the rod,

n + TSin$\theta$ = W

n + (58.23)Sin(38.5) = 72.5

n = 36.25 N

Vertical normal forces exerted by the support = 36.25 N

Tension in the cable = 58.23 N