|
martindcx1e
|
02-21-2008, 01:58 PM
Post subject: A force/friction physics problem
|
#1 (permalink)
|
|
4-of-a-Kind
Join Date: Mar 2005
Posts: 3,614
|
|
Can anyone help with this?
A 420 g block is at rest on a horizontal surface. The coefficient of static friction between the block and the surface is 0.800, and the coefficient of kinetic friction is 0.230. A force of magnitude P pushes the block forward and downward. Assume the force is applied at an angle of 23.0° below the horizontal.
(a) Assuming P is large enough to make the block move, find the acceleration of the block as a function of P. (Use P as necessary and round numbers to the third decimal place.)
a(P) = ?
(b) If P = 4.12 N, find the acceleration and the friction force exerted on the block.
(c) If P = 8.24 N, find the acceleration and the friction force exerted on the block.
For A I have to set the net force equal to m*a right? If so then I got n = mg + Psin(-23) for the vertical forces and for the horizontal forces I'm not sure if I'm supposed to use both the force of static friction and the force of kinetic friction in my equation or just one or the other.
|
|
Wikipedia is the best thing ever. Anyone in the world can write anything they want about any subject. So you know you are getting the best possible information.
|
Play for FREE and practice your game at...
Join the FTR Poker Forum to disable these banners and start posting!
|
|
a500lbgorilla
|
|
02-21-2008, 03:45 PM
Post subject: Re: A force/friction physics problem
|
#2 (permalink)
|
|
JESUS TAKE THE KEYBOARD
Join Date: Sep 2004
Location: This room is a good place to be
Posts: 8,379
|
|
Quote:
|
Originally Posted by martindcx1e
Can anyone help with this?
A 420 g block is at rest on a horizontal surface. The coefficient of static friction between the block and the surface is 0.800, and the coefficient of kinetic friction is 0.230. A force of magnitude P pushes the block forward and downward. Assume the force is applied at an angle of 23.0° below the horizontal.
(a) Assuming P is large enough to make the block move, find the acceleration of the block as a function of P. (Use P as necessary and round numbers to the third decimal place.)
a(P) = ?
In order to make the block move from rest, the force has to exceed the static friction force. Static friction comes in at Fs = (.42*9.81 + Psin(23))*.8.
When this force is exceeded, the firction force drops to Fk = (.42*9.81 + Psin(23))*.23 When this occurs, the sum of the forces in the X shape up to be .42* a = P*cos(23) - Fk. Solve for a
(b) If P = 4.12 N, find the acceleration and the friction force exerted on the block.
First check to see if the P is large enough to overcome Fs. Fs = 4.584. Px = 3.792 N. So the acceleration = 0, and equal and opposite forces says the force of friction = 3.792 N.
(c) If P = 8.24 N, find the acceleration and the friction force exerted on the block.
Fs = 5.871 N, Px = 7.585 N. Force of friction = Fk = 1.688 N. a = 14.04 m/s/s. Might wanna check my math
For A I have to set the net force equal to m*a right? If so then I got n = mg + Psin(-23) for the vertical forces and for the horizontal forces I'm not sure if I'm supposed to use both the force of static friction and the force of kinetic friction in my equation or just one or the other.
|
The force of static friction only comes into play when the block is not moving and is usually much higher than the force of kinetic friction. Until the force applies exceeds the static frictional force in X, the block will not move. Once it starts moving, it will only resist by the force of kinetic friction.
|
Smithers, use the amnesia ray.
You mean the revolver, sir?
Precisely.
|
|
kingnat
|
|
Full House
Join Date: Jun 2005
Posts: 827
|
|
The notion that the frictional force in part b) is actually equal to the horizontal component of the pushing force is something that people get wrong often... rilla did it exactly right...
Checking that the pushing force (x-comp only) is less than the maximum static friction force tells us that it doesn't move, but that means the ACTUAL static frictional force is simply equal to the x-component of the pushing. People often will incorrectly answer that the static frictional force IS the maximum static frictional force. just thought i'd mention that in case you don't recognize the distinction in 'Rilla's solution.
|
|
So you click their picture and then you get their money?
|
|
martindcx1e
|
|
4-of-a-Kind
Join Date: Mar 2005
Posts: 3,614
|
|
geez i'm an idiot. i totally missed that in part A it said to assume that P makes the block move. thanks for the help though rilla.
|
|
Wikipedia is the best thing ever. Anyone in the world can write anything they want about any subject. So you know you are getting the best possible information.
|
|
a500lbgorilla
|
|
JESUS TAKE THE KEYBOARD
Join Date: Sep 2004
Location: This room is a good place to be
Posts: 8,379
|
|
my pleasure
|
Smithers, use the amnesia ray.
You mean the revolver, sir?
Precisely.
|
|
wufwugy
|
|
4-of-a-Kind
Join Date: May 2006
Posts: 1,660
|
|
physics is a myth
|
|
|
|
Lukie
|
|
4-of-a-Kind
Join Date: Jul 2005
Location: back with a vengeance
Posts: 3,307
|
|
|
|
|
|
martindcx1e
|
|
4-of-a-Kind
Join Date: Mar 2005
Posts: 3,614
|
|
Quote:
|
Originally Posted by Lukie
|
LOL that rocks
|
|
Wikipedia is the best thing ever. Anyone in the world can write anything they want about any subject. So you know you are getting the best possible information.
|
|
swiggidy
|
|
4-of-a-Kind
Join Date: Sep 2005
Location: Waiting in the shadows ...
Posts: 3,777
|
|
Quote:
|
Originally Posted by Lukie
|
FYP
|
|
(\__/)
(='.'=)
(")_(")
|
|
martindcx1e
|
|
4-of-a-Kind
Join Date: Mar 2005
Posts: 3,614
|
|
Quote:
|
Originally Posted by swiggidy
Quote:
|
Originally Posted by Lukie
|
FYP |
forgive him he's a bit slow
|
|
Wikipedia is the best thing ever. Anyone in the world can write anything they want about any subject. So you know you are getting the best possible information.
|
|
Lukie
|
|
4-of-a-Kind
Join Date: Jul 2005
Location: back with a vengeance
Posts: 3,307
|
|
the town is back THAT way!
|
|
|
|
Galapogos
|
|
4-of-a-Kind
Join Date: Jun 2005
Location: The Loser's Lounge
Posts: 2,322
|
|
Quote:
|
Originally Posted by Lukie
the town is back THAT way!
|
Classic.
|
Quote:
|
Originally Posted by sauce123
I don't get why you insist on stacking off with like jack high all the time.
|
|
English
Deutsch
Español
Français
Italiano
Nederlands
Pусский
Svenska
05-26-2012, 03:08 PM Australia Legalized Online Poker coming up in next 6 to 12 Months
Linear Mode
