Moment about a point problems discussion

Question 1: A vertical force of known magnitude 1000 N is needed to remove nail C from the board.
The time at which this nail is about to move up, determine (a) the moment around B of the force exerted on the nail, (b) the amount of force P that creates equal moment about B if angle ⦤ȧ = 10°, (c) the smallest force P that creates the same moment about B.

Solution:

Question 2: For the rectangular lamina shown below, find the total moment of the forces acting, about the corner marked O.

Solution:

Force	Moment (Nm)
5N at O	50=0
8 N	−81.2=−9.6 (-ve sign is because of )
7 N	70=0
6 N	−60.5=−3
5 N	51.2=6
4 N	40.5=2
Total Moment is:	0−9.6+0−3+6+2=−4.6 N.m.

Question 3: 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 maximum weight or mass that can be placed on either end of the beam to keep it in equilibrium.

Solution:

The diagram shows the forces acting on the beam.

Taking moments about the point where R1 acts gives:

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

Taking moments about the point where R2 acts gives:

2R1=0.5490 R1=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

Consider the greatest mass which can be placed on left hand at the end of the beam. Diagram shown below is displaying 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=1.5490 m=9.81.5490=75 kg  

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

2mg=0.5490 m=2g0.5490=12.5 kg  

Hence the greatest mass that can be placed at either end of the beam is 12.5 kg.