Question

A 10.0 metre long horizontal steel bar with a mass of 46 kg, is pivoted at a vertical wall and supported by a cable that makes an angle of 25 degrees to the bar. The cable is attached to the end of the bar, at the opposite end of the bar to the wall, and is anchored on the wall above the pivot point.

(a) Sketch a force diagram for this situation, and define all the forces acting on the bar You can also include a sketch of the actual physical layout.

(b) What is the tension in the cable? Upload this answer below during your online session.

(c) What is the force (in newton) exerted by the wall against the bar. Include magnitude and direction of this force in your fully worked solution.

Answer 1

Figure shows the forces acting on the bar that is supported by a cable anchored to wall.

W = mg is the weight , i.e. gravitational force acting on bar, T is Tension force acting on the cable,

R is reaction force from wall , acting on bar at contact point.

Tension force is resolved into horizontal and vertical component as shown in figure.

At equlibrium, we have ,   ( m g ) = T sin25

Hence Tension force T = ( m g ) / sin25 = ( 46 9.8 ) / sin25 = 1067 N

Tension force is acting along the direction of cable as shown in figure.

Reaction force R = T cos25 = (1067 cos25 ) = 967 N

Reaction force from wall is acting at the contact point of wall and bar . Direction of reaction force is parallel to bar