14. Moments and equilibrium

From Mechanics

Revision as of 10:33, 23 September 2009

14. Moments and Equilibrium

Key Points

In equilibrium: • The resultant force on a body must be zero. and • The resultant moment on a body must be zero.

Example 14.1

A uniform beam has length 8 m and mass 60 kg. It is suspended by two ropes, as shown in the diagram below.

Find the tension in each rope.

Solution

The diagram shows the forces acting on the beam.

Take moments about the point where T1 acts to give:

5T2=3588T2=53588=352.8=353 N (to 3sf)

Take moments about the point where T2 acts to give:

5T1=2588T1=52588=235.2=235 N (to 3sf)

Finally for vertical equilibrium we require T1+T2=588 , which can be used to check the tensions. In is the case we have:

352.8+235.2=588

Example 14.2

A beam, of mass 50 kg and length 5 m, rests on two supports as shown in the diagram. Find the magnitude of the reaction force exerted by each support.

Find the maximum mass that could be placed at either end of the beam if it is to remain in equilibrium.

Solution

The diagram shows the forces acting on the beam.

Taking moments about the point where R1 acts gives:

2R2=1.5490R2=21.5490=367.5=368 N (to 3sf)

Taking moments about the point where R2 acts gives:

2R1=0.5490R1=20.5490=122.5=123 N (to 3sf)

For vertical equilibrium we require R1+R2=490 ,

which can be used to check the tensions. In is the case we have:

367.5+122.5=490

First consider the greatest mass that can be placed at the left hand end of the beam. The diagram below shows the extra force that must now be considered. When the maximum possible mass is used, R2=0 .

Taking moments about the point where R1 acts gives:

1mg=1.5490m=9.81.5490=75 kg

Similarly for a mass placed at the right hand end of the beam:

2mg=0.5490m=2g0.5490=12.5 kg

Hence the greatest mass that can be placed at either end of the beam is 12\textrm{.}5 kg.

Example 14.3

A ladder, of length 3 m and mass 20 kg, leans against a smooth, vertical wall so that the angle between the horizontal ground and the ladder is 60. (a) Find the magnitude of the friction and normal reaction forces that act on the ladder, if it is in equilibrium. (b) Find the minimum value of the coefficient of friction between the ladder and the ground.

Solution

The diagram shows the forces acting on the ladder.

(a) Considering the horizontal forces gives:

F=S

Considering the vertical forces gives:

R=196

Taking moment about the base of the ladder gives:

1961.5cos60=S3sin60S=3sin601961.5cos60=2sin60196cos60=1962tan60=56.6 N (to 3 sf)

But since F=S , the friction force has magnitude 56\textrm{.}6 N.

(b)

Using the friction inequality, FR

gives:

1962tan6019612tan600.289 (to 3sf)