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Part of the floor of a workshop is inclined at 10 degrees to the horizontal. This is to allow the safe storage of steel cylinders. A cylinder of mass 7000 kg is stored as shown in the diagram (c the link below). The magnitudes of the forces exerted on the cylinder by the wall and the floor are Rw and Rf newtons respectively. Find the values of Rw and Rf.

If the sloping part of the floor was inclined at alpha degrees (0 < alpha < 45) instead of at 10 degrees, show that

a) Rw < 70000< Rf, whatever the value of alpha.

b) Rw and Rf both increase with alpha.

http://www.flickr.com/photos/8625954...in/photostream

i did this: R (II to slope) : Rwcos10 = 70000sin10 ---- Rw = 12342.9 N R (perpendicular to slope) : Rf = 70000cos10 + 12342.9sin10 ----- Rf = 71079.9 N

I am able to get Rw and Rf on the cylinder. But I don't know how to approach mathematically expressing the second part of the question (the proof part). Parts (a) and (b) of the second part. Could you please help?

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1 Answer

The link in your question appears to be broken, so I haven't seen the picture. However, based on what I imagine the picture looks like, my suggestion as to how to approach the problem is as follows:

a) Double-check your work for the forces exerted by the wall and the floor. Remember that the force equations are:
Parallel Force = $mgsin(θ)$
Perpendicular Force = $mgcos(θ)$

b) As the value of $\alpha$ increases from 0 to 45 degrees, we can see that $cos(\alpha)$ will decrease from 1 to $\frac{1}{\sqrt{2}}$ while $sin(\alpha)$ will increase from 0 to $\frac{1}{\sqrt{2}}$. Watch what increasing $\alpha$ does to the values for force in each direction and comment on the extreme values (where $\alpha$ = 0 and $\alpha$ = 45).

Hope this helps, and if not be sure to comment back once the link is up and I'll take a look at the specifics of the problem!

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