[Chikis (OP)]

A uniform beam  6.0 metres long and weighing 4 kg rest on supports at P and Q placed left and right 1.0 metre from each end of the beam. Weights of mass 10 kg and 8 kg are placed near P and Q respectively one  each end of the beam. Calculate the reactions at P and Q?

After drawing the diagram for the whole set up,  assuming that clockwise moment = anticlockwise moment, I then took the moment about Q. Where
Rp = reaction at p
4*Rp = (2+1)10 + (2*4)
>
Rp = 38/4 * 10 = N95

Am having problem getting the reaction Rq at Q. This is what I did:

Am having problem getting the reaction Rq at Q. This is what I did:

Taking the moment about pivot P.
Clock wise moment = anti clockwise momen.

(2*4) + (3* = Rq*5
>
Qr = 32/5 * 10  = N 6.4
 Pleaas help me. What am suppose to do that am not doing?

[alancalverd]

I put P and the larger mass on the left.

Clockwise moment about Q is 8 *1 (mass on the right x lever arm) + Rp*4 (reaction at P x lever arm)
Anticlockwise moment = 10*5 (mass at P) + 4*2 (mass of beam acting through its center of mass)

So 8 + 4Rp = 50 + 8 and Rp = 12.5

Since the total downward force is 22, Rq must be 9.5, but if you want to check your arithmetic

Clockwise moment about P = 4*2 (mass of beam) + 8*5 (mass at Q)

Anticlockwise = 10*1 (mass at P) + Rq*4

So 10 + 4Rq = 8 + 40 and Rq = 9.5

I think you confused your P's and Q's.

[Chikis (OP)]

I think you confused your P's and Q's.

How? I took the left and right ends of the beam P and Q where the weights are attached to be my left and right respectively. Should I have assumed that the beam is a person standing facing me so that the left and right where the weight are attached become my own?

[alancalverd]

Makes no difference. I have no idea where the numbers came from in your working, or what N stands for.

Just draw the picture and write down each moment as force x distance, then add them up.

Remember that  the weight of the beam acts through its center, and the reaction force is upwards.

[Chikis (OP)]

Makes no difference. I have no idea where the numbers came from in your working, or what N stands for.

Just draw the picture and write down each moment as force x distance, then add them up.

Remember that  the weight of the beam acts through its center, and the reaction force is upwards.

[Chikis (OP)]

Makes no difference. I have no idea where the numbers came from in your working, or what N stands for.

Just draw the picture and write down each moment as force x distance, then add them up.

Remember that  the weight of the beam acts through its center, and the reaction force is upwards.

I can not draw any picture more than the first I had drawn. Is there any way you can show me the diagram that you have drawn for the problem. Then it will be easy for me to understand the calculation you have done.

[alancalverd]

http://www.thenakedscientists.com/forum/diagrams/sketch691_0.gif

But please tell me what N means in your calculation!

[Chikis (OP)]

But please tell me what N means in your calculation!

N is Newton, the unit of the reaction forces.

[Chikis (OP)]

http://www.thenakedscientists.com/forum/diagrams/sketch691_0.gif [nofollow]

In the diagram, you never showed the pivots P and Q below the reactions forces. But even though you did not  show it, I already know they are there because without it they can't be reaction forces at P and Q.

[alancalverd]

Sorry, I couldn't find a neat triangle to illustrate the support points on the drawing program! Anyway I think the diagram is pretty obvious.

You (or whoever posed the question in the first place) clearly haven't fully grasped the notion of mass and weight. The question talks rather loosely about weights in kilograms. Weight is a force, so the reaction force is calculated in the same units and there is an inherent assumption that we are talking about  "kilogram force", i.e. the gravitational force mg on a 1 kg mass m. Since g was not specified, we can't assume a value, so we can't report the answer in newtons. It would be a lot less on the moon, or even on top of a mountain on earth, whereas the kilogram forces would be exactly as calculated.