Hey, so this is probably very easy but I'm not seeing how I'm solving this without using the friction coefficient (might be called friction factor).
I did an experience on class the other day and now I have to answer some questions.
One of those is, I need to theoretically calculate the acceleration of a block of wood and its friction's force. Here's a picture to ilustrate the situation
http://i.imgur.com/PFldzIE.png
I have the mass of both blocks, which are A (0,12 kg) and B (0,04 kg) but everytime I try and solve it, I get 2 unknown values which are both friction coefficient and aceleration.
I use the formulas Fx = T - Ff <=> m x a= mb x g - uc x Nr <=> m x a = mb x g - uc x
After that I can't do it because I don't know a or uc.
uc means friction coefficient
Nr means normal reaction
Below block A, there's a surface, and B is suspended.
EDIT: I'm not downfragging everyone btw... I guess I'm coming to the conclusion my teacher isn't very smart or something
Hey, so this is probably very easy but I'm not seeing how I'm solving this without using the friction coefficient (might be called friction factor).
I did an experience on class the other day and now I have to answer some questions.
One of those is, I need to theoretically calculate the acceleration of a block of wood and its friction's force. Here's a picture to ilustrate the situation
[IMG]http://i.imgur.com/PFldzIE.png[/IMG]
I have the mass of both blocks, which are A (0,12 kg) and B (0,04 kg) but everytime I try and solve it, I get 2 unknown values which are both friction coefficient and aceleration.
I use the formulas Fx = T - Ff <=> m x a= mb x g - uc x Nr <=> m x a = mb x g - uc x
After that I can't do it because I don't know a or uc.
uc means friction coefficient
Nr means normal reaction
Below block A, there's a surface, and B is suspended.
EDIT: I'm not downfragging everyone btw... I guess I'm coming to the conclusion my teacher isn't very smart or something
okay. so the weight of block a to block b is 1/3. which should be the friction coefficient. http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html
so then the acceleration should be 2.205.
http://hyperphysics.phy-astr.gsu.edu/hbase/hpul2.html#c1
while i'm literally taking high-school physics right now, i could be completely wrong. as most of my test scores can tell you.
EDIT: shit. i tried.
okay. so the weight of block a to block b is 1/3. which [i]should[/i] be the friction coefficient. http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html
so then the acceleration should be 2.205.
http://hyperphysics.phy-astr.gsu.edu/hbase/hpul2.html#c1
while i'm literally taking high-school physics right now, i could be completely wrong. as most of my test scores can tell you.
EDIT: shit. i tried.
Manko, you are right that you are missing information. You can't solve for a without knowing uc. Your equations look fine to me, too.
To see this intuitively, consider the two extreme cases -- with sufficiently high friction, the friction on block A could support the weight of block B (imagine gluing A to the table), and it doesn't even have to move. With zero friction, block A is going to accelerate as if you were pulling it with a force equal to the weight of block B. Since you have different accelerations with different friction coefficients, you can't know a without knowing something more about uc.
Manko, you are right that you are missing information. You can't solve for a without knowing uc. Your equations look fine to me, too.
To see this intuitively, consider the two extreme cases -- with sufficiently high friction, the friction on block A could support the weight of block B (imagine gluing A to the table), and it doesn't even have to move. With zero friction, block A is going to accelerate as if you were pulling it with a force equal to the weight of block B. Since you have different accelerations with different friction coefficients, you can't know a without knowing something more about uc.
impossible to get acceleration as a number, it will need to be in terms of T (tension in the wire) or μ (coef. of friction). if you are assuming g = 9.81 then a = 2.4525 - 7.3575μ
impossible to get acceleration as a number, it will need to be in terms of T (tension in the wire) or μ (coef. of friction). if you are assuming g = 9.81 then [b]a = 2.4525 - 7.3575μ[/b]
Draw free body diagrams.
Let Ft be the tension in the string, A be the mass of block A, and B be the mass of block B, and little a is acceleration in the system.
So for block A, the net force acting on it is A * a = Ft - uc * g * A. (right is positive)
For block B, the net force acting on it is B * a = g * B - Ft (down is positive)
So we can write Ft = g * B - B * a.
Plugging this in we get
A * a = g * B - B * a - uc * g * A
A * a + B * a = g * B - uc * g * A
a = g * (B - uc * A)/(A + B)
= 9.8 * (.04 - .12uc)/(.16)
= what falcon got
Draw free body diagrams.
Let Ft be the tension in the string, A be the mass of block A, and B be the mass of block B, and little a is acceleration in the system.
So for block A, the net force acting on it is A * a = Ft - uc * g * A. (right is positive)
For block B, the net force acting on it is B * a = g * B - Ft (down is positive)
So we can write Ft = g * B - B * a.
Plugging this in we get
A * a = g * B - B * a - uc * g * A
A * a + B * a = g * B - uc * g * A
a = g * (B - uc * A)/(A + B)
= 9.8 * (.04 - .12uc)/(.16)
= what falcon got
Falcon0408impossible to get acceleration as a number, it will need to be in terms of T (tension in the wire) or μ (coef. of friction). if you are assuming g = 9.81 then a = 2.4525 - 7.3575μ
Can you tell me how you got there?
EDIT: -dan just got it, thanks. I'll tell my teacher I couldn't find the coefficient on the books and that she's retarded.
[quote=Falcon0408]impossible to get acceleration as a number, it will need to be in terms of T (tension in the wire) or μ (coef. of friction). if you are assuming g = 9.81 then [b]a = 2.4525 - 7.3575μ[/b][/quote]
Can you tell me how you got there?
EDIT: -dan just got it, thanks. I'll tell my teacher I couldn't find the coefficient on the books and that she's retarded.
I PMed -dan about a question I have, but he doesn't seem to be responding if someone can answer it I'll appreciate :)
when he goes from A * a + B * a = g * B - uc * g * A <=> a = g * (B - uc * A)/(A + B) shouldn't there be 2a instead of only a, both A and B go to the other side dividing while both a's stay...
I PMed -dan about a question I have, but he doesn't seem to be responding if someone can answer it I'll appreciate :)
when he goes from A * a + B * a = g * B - uc * g * A <=> a = g * (B - uc * A)/(A + B) shouldn't there be 2a instead of only a, both A and B go to the other side dividing while both a's stay...
no, you distribute the a out so it's a(A+B)
no, you distribute the a out so it's a(A+B)
Looks like you would solve it using a system of equations.
Looks like you would solve it using a system of equations.