Answer to Question #271550 in Mechanics | Relativity for Sonia Charles

Question

This question is about forces and moments of forces.

a. (I) Define the moment of a force about a point

(II) State the law of moments as a condition for equilibrium

b. A crane has a jib (AC) of mass 2500kg pivoted at A. 

The jib is supported by a horizontal cable BC. The centre of mass of the jib its at D. The dimensions are as shown on the diagram.

(I) Make a sketch copy of the diagram and on it draw three arrows to represent the three forces acting on the jib.

(ii) Calculate the tension in the cable BC.

C. The end of the jib C is raised by shortening BC. Explain whether the tension in BC will increase or decrease. 

Answer 1

a)

i)

moment of a force about a point: the product of the distance from the point to the point of application of the force and the component of the force perpendicular to the line of the distance.

ii)

law of moments as a condition for equilibrium: when a body is balanced, the total clockwise moment about a point equals the total anticlockwise moment about the same point.

b)

i)

m is mass of jib

N is reaction

T tension of cable

ii)

$T|AC|sin30\degree=mg|AD|cos30\degree$

$T=\frac{mg|AD|cos30\degree}{|AC|sin30\degree}=\frac{2500\cdot 4gcos30\degree}{8sin30\degree}=2165$ N

c)

let $\alpha$ be angle between jib and horizontal

When the end of the jib C is raised, $cot\alpha$ decreases. So, the tension in BC will decrease.