Question

A uniform bar is attached to a wall by a frictionless hinge at its lower end. The bar is held at an angle of 30 degrees above the horizontal by a cable that is attached to the upper end of the bar. The cable makes an angle of 53 degrees with the bar. The tension in the cable is 80 N. A) What is the mass of the bar? B) What are the magnitude and direction of the resultant force that the hinge exerts on the bar? Specify the direction by giving the angle that the force makes with the horizontal.

Answer 1

let mass of the bar be m

then

tension in the cable = T = 80 N

then form moment balance about the hinge we get

T*sin(53)*L = mgcos(30)L/2

here L is length of the bar

m = 2T*sin(53)/g*cos(30)

m = 15.0407315542 kg

B. let the reacitons of the hinge on the bar be Rx, Ry

then from force balance

Tcos(53 + 30) = Rx = 9.749547472411798489031513538522 N

Tsin(53 + 30) + Ry = mg

Ry = 68.14588441539623720159 N

hence

|R| = sqroot(Rx^2 + Ry^2) = 68.839779478 N

resultant direction = phi wrt horizontal

tan(phi) = Ry/Rx

hence

phi = 81.85801548532376 deg to the horizontal