# Mechanics

• Dec 27th 2010, 09:49 AM
BabyMilo
Mechanics
http://www.mathhelpforum.com/math-he...s-untitled.jpg

Which is the right answer of R?

Rcos27-85g=0
R= 85g/cos27
R= 935N

or

R-85gcos27=0
R=85gcos27
R= 742N

Im bit confused. In M1, I was told to use the Second method, which is right for M1.
But Im doing M2 atm, and the solution uses the first method.

Can someone explain?

• Dec 27th 2010, 09:53 AM
skeeter
$R = mg\cos{\theta}$
• Dec 27th 2010, 09:55 AM
BabyMilo
So the second one is right?
• Dec 27th 2010, 10:40 AM
skeeter
Quote:

Originally Posted by BabyMilo
So the second one is right?

if $85g$ is the weight of the mass, then $R = 85g\cos(27^\circ)$
• Dec 27th 2010, 10:42 AM
snowtea
If you ever get confused, just remember that the normal force on a ramp cannot exceed the weight (unless something else is pushing it against the ramp).
• Dec 27th 2010, 11:09 AM
BabyMilo
Quote:

Originally Posted by skeeter
if $85g$ is the weight of the mass, then $R = 85g\cos(27^\circ)$

what if there is a horizontal force of mw^2r?

would R still be 85gcos27 ?

thanks
• Dec 27th 2010, 11:19 AM
snowtea
Quote:

Originally Posted by BabyMilo
what if there is a horizontal force of mw^2r?

would R still be 85gcos27 ?

thanks

You are talking about centripedal force right? In that case 85g/cos27.

Remember, if the block is sliding down the ramp, the force is pointing down (parallel) to the ramp. In this case, the force components perpendicular to the ramp should sum to zero. If you draw the free body diagram, you get R = 85gcos(27).

If it is the angle of a ramp in a turn, then the force is horizontal (pointed directly towards the center of the circle). In this case, the vertical force components should sum to zero. If you draw the free body diagram, you get R = 85g/cos(27).

Also, usually R means radius and not normal force. I would suggest changing the notation to avoid confusion.
• Dec 27th 2010, 12:00 PM
BabyMilo