Elavating Ramp
A block is at rest on an inclined plane whose elevation can be varied. The coefficient of static friction is = 0.37, and the coefficient of kinetic friction is = 0.15. The angle of elevation is increased slowly from the horizontal.
At what value of does the block begin to slide (in degrees)?
2.031×101
What is the acceleration of the block?
2.022 m/s^2
At what value of does the block begin to slide (in degrees)?
2.031×101
What is the acceleration of the block?
2.022 m/s^2
Discussion
help Anonymous 1 (Thu Jan 21 12:26:30 am 2010 (EST))
I am doing this problem EXACTLY how we were shown in class
and i am getting the answer wrong.. can someone explain this?
i am using Theta= tan^-1 (Mu s)
NEW
Anonymous 2 (Thu Jan 21 03:26:32 pm 2010 (EST))
any help on part 2?
NEW
since the static friction equals tan(theta), you have to use the inverse of tangent
to find the angle. Hope that helps.
NEW
for part 2
newton 2 says EF=m*a
you can ignore forces on the y axis
forces on x: forward force -Friction
EF= m*g sin(theta) - muK * N
EF= m*g sin(theta) - muK * m*g*cos(theta)= m*a
solve for a
NEW
Anonymous 4 (Thu Jan 21 07:54:52 pm 2010 (EST))
for part 2 what is m?
NEW
The mass isn't provided. You have to use the equation that
Dr. Ing gave us last lecture which was:
a = g(sin theta) - (mu k)*(cos theta)
NEW
Anonymous 4 (Fri Jan 22 01:45:19 pm 2010 (EST))
I am using the eqn 9.8(sin theta i got for part 1)- mu K
(cos theta i got for part one) and im not getting it right?
Any help? I may be putting in the wrong units?
NEW
Re: help Anonymous 6 (Fri Jan 22 03:19:50 pm 2010 (EST))
Just make sure that your calculator is in degrees rather
than radians and it should work out.
NEW
I'm having the same problem,
I have a=9.8(sin 19.3)-.16(cos 19.3)
So my a is 3.088 m/s^2
But it's incorrect according to Lon Capa. Anyone have a clue
where I'm messing up?
NEW
the equation is a=g((sin theta)-(mu k)(cos theta))..make
sure g isn't just multiplied by the sin theta and the
answer should be correct
NEW
Thank you, that was exactly the problem!
I am doing this problem EXACTLY how we were shown in class
and i am getting the answer wrong.. can someone explain this?
i am using Theta= tan^-1 (Mu s)
NEW
Anonymous 2 (Thu Jan 21 03:26:32 pm 2010 (EST))
any help on part 2?
NEW
since the static friction equals tan(theta), you have to use the inverse of tangent
to find the angle. Hope that helps.
NEW
for part 2
newton 2 says EF=m*a
you can ignore forces on the y axis
forces on x: forward force -Friction
EF= m*g sin(theta) - muK * N
EF= m*g sin(theta) - muK * m*g*cos(theta)= m*a
solve for a
NEW
Anonymous 4 (Thu Jan 21 07:54:52 pm 2010 (EST))
for part 2 what is m?
NEW
The mass isn't provided. You have to use the equation that
Dr. Ing gave us last lecture which was:
a = g(sin theta) - (mu k)*(cos theta)
NEW
Anonymous 4 (Fri Jan 22 01:45:19 pm 2010 (EST))
I am using the eqn 9.8(sin theta i got for part 1)- mu K
(cos theta i got for part one) and im not getting it right?
Any help? I may be putting in the wrong units?
NEW
Re: help Anonymous 6 (Fri Jan 22 03:19:50 pm 2010 (EST))
Just make sure that your calculator is in degrees rather
than radians and it should work out.
NEW
I'm having the same problem,
I have a=9.8(sin 19.3)-.16(cos 19.3)
So my a is 3.088 m/s^2
But it's incorrect according to Lon Capa. Anyone have a clue
where I'm messing up?
NEW
the equation is a=g((sin theta)-(mu k)(cos theta))..make
sure g isn't just multiplied by the sin theta and the
answer should be correct
NEW
Thank you, that was exactly the problem!