M1 - Friction & Coefficient of Friction Help?

Hello! I was wondering if anyone could give me a bit of help 'cause I'm a bit rubbish at Mechanics and I've got stuck?

The question is:

The diagram shows a particle 'P' of mass 4kg on a smooth plane inclined at 15 degrees to the horizontal. 'P' is held in equilibrium by a horizontal force, 'F'.

Show that the normal reaction exerted by the plane on 'P' is 40.6 N correct to 3 significant figures.

________________________

See, I thought it'd just be 4 x 9.8 x cos15, but that gives me 37.9. :/

Thanks in advance!

Reply 1

made_of_fail

what exactly are you stuck on?

and no coefficient provided?

Reply 2

Ziggy_Stardust

I'm stuck on showing that the normal is 40.3, like the question says it is. I'm only getting 37.9. And nope - no coefficient.

Reply 3

made_of_fail

woops, didn't read the question properly.

im getting the same as you.

resolving parallel to the plane means that the only forces involved are the weight and the normal reaction, so:

R=4gcos15 as you said.

also, according to the question costheta=40.1/4g=1.02 !?

Reply 4

Ziggy_Stardust

Yeah, that's what I was thinking. Oh well. Thanks for looking anyway! :smile:

Reply 5

manderton

made_of_fail

R=4gcos15

That is incorrect.

Resolving forces vertically gives:

RCos15 = (4*9.8)

therefore:

R = \frac {(4*9.8)}{cos15}

which equals 40.6N

Reply 6

Ziggy_Stardust

Ahhh. I see now. Thank you!

Reply 7

made_of_fail

ah i see i didnt look at the diagram. was thinking F was parallel to the slope. really!

Reply 8

manderton

it will become easier as u progress especially if u go onto m2.

pm if need help with anything else - im revising m1 atm to help with m2 so will be willing to help with anything :smile:

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