# Thread: resolving forces, should be simple

1. ## resolving forces, should be simple

Two cylinders of equal radii each have weight W. They are at rest on a smooth slope of inclination A° to the horizontal, the lower of the two cylinders being in contact with a smooth vertical wall as shown in the diagram. Find, giving your answer in terms of W and A°, the magnitude of the force exerted by
a. the slope on the upper cylinder
b. the lower cylinder on the upper cylinder
c. the upper cylinder on the lower cylinder
d. the wall on the lower cylinder
e. the slope on the lower cylinder

OK so the first three are pretty obvious, a.= WcosA°, b.=c.=WsinA°
However I’m having trouble with d.

So d.=(normal contact force exerted on lower cylinder by slope resolved for the horizontal)+(normal contact force exerted on lower cylinder by upper cylinder resolved for the horizontal) = ((WcosA° resolved for the horizontal) + (WsinA° resolved for the horizontal)) = WcosA°sinA° + WsinA°cosA°=2WsinA°cosA°
However, the correct answer is 2WtanA°; where did I go wrong?

2. Originally Posted by furor celtica
Two cylinders of equal radii each have weight W. They are at rest on a smooth slope of inclination A° to the horizontal, the lower of the two cylinders being in contact with a smooth vertical wall as shown in the diagram. Find, giving your answer in terms of W and A°, the magnitude of the force exerted by
a. the slope on the upper cylinder
b. the lower cylinder on the upper cylinder
c. the upper cylinder on the lower cylinder
d. the wall on the lower cylinder
e. the slope on the lower cylinder

OK so the first three are pretty obvious, a.= WcosA°, b.=c.=WsinA°
However I’m having trouble with d.

So d.=(normal contact force exerted on lower cylinder by slope resolved for the horizontal)+(normal contact force exerted on lower cylinder by upper cylinder resolved for the horizontal) = ((WcosA° resolved for the horizontal) + (WsinA° resolved for the horizontal)) = WcosA°sinA° + WsinA°cosA°=2WsinA°cosA°
However, the correct answer is 2WtanA°; where did I go wrong?
The bits in red are where you are going wrong. It is not correct to assume that the normal force exerted by the slope is the same on the lower cylinder as it is on the upper cylinder. The horizontal force H exerted by the wall has a component normal to the slope, which has to be balanced by an increased normal force of the slope on the cylinder.

The best way to deal with part d. is to resolve the forces on the lower cylinder in the direction of the slope. You will then have a component HcosA° of the horizontal force, balanced by a component WsinA° of the weight of the lower cylinder plus another WsinA° given by the normal contact force exerted on lower cylinder by upper cylinder.

3. thanks for your time man

ok so i get it now, i forgot the force exerted by the wall on the lower cylinder in calculating the normal reaction of the slope on this cylinder
so it gives this
H= WsinAcosA + (WcosA + HsinA)sinA
H-H(sinA)^2= 2WsinAcosA
H=(2WsinAcosA)/(1-(sinA)^2)=2WtanA
but this explanation require the identity (cosA)^2=(1-(sinA)^2), and i'm not sure i'm supposed to use this identity at this stage in my course. is there an alternate method of calculation to find the result 2WtanA?

4. Or maybe another way to solve d, other than the explanation I posted on PHF, is:

Find the force P which maintains the cylinder stationary on the slope (assume no friction)

Then, you apply the same principle with the two cylinders.