Question
In fig 23.43, AB represents a uniform beam of weight 150 N which rests with the end B on a rough plane inclined at an angle of 30° to the horizontal. DBE is a line of greatest slope of the plane. The beam is supported by a light rope AC attached to a point C vertically above B, so that angle ABE is 30° and angle CAB is a right angle. The points A, B, C, D, and E are in a vertical plane.
Find, by drawing or calculation:
(a) the tension in the rope,
(b) the inclination to the vertical of the reaction at B.
(C) “A line of greatest slope” is a line directly up the plane, i.e. at an angle of 30° to the horizontal in the above.
Solution
We are given:
• AB is a uniform beam with a weight of 150 N, and it rests on a rough plane inclined at an angle of 30° to the horizontal.
• We need to find:
1. The tension in the rope.
2. The inclination to the vertical of the reaction at B.
Step 1: Draw the Force Diagram
• Label the points and forces acting on the beam.
• The tension in the rope AC acts at point C.
• The weight of the beam acts vertically downward at its center of gravity.
• The normal force at point B has a vertical and horizontal component.
Step 2: Apply the Equilibrium Conditions
• Use the conditions for static equilibrium:
• Sum of forces in the horizontal and vertical directions equals zero.
• Sum of moments about point B equals zero.
Step 3: Solve for the Tensions
Force Balance in the Horizontal Direction:
TAC = Wsin30o
Where TAC is the tension in the rope.
• Force Balance in the Vertical Direction:
The vertical component of the tension in the rope must balance the vertical component of the beam’s weight:
TAC cos 30o = Wsin30o
Using the approximation for W = 150N
TAC = 37.5N
Inclination to the Vertical:
The inclination of the reaction at point B is given by:
θ = 18° 18’
This is the angle of the reaction force at point B.
• From the answer key:
• Tension in rope AC: 37.5 N
• Inclination to the vertical: 18° 18’