Block Tension and Friction Question

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The discussion revolves around calculating the acceleration of a 2.0 kg block on a table with a coefficient of kinetic friction of 0.350. Participants are attempting to solve the problem using free body diagrams and Newton's laws, but are encountering issues with their calculations and assumptions about tension in the system. Several equations are proposed, but confusion arises regarding the correct application of forces and the direction of acceleration. Ultimately, the participants are encouraged to combine their equations correctly to arrive at the right answer, with one user suggesting that the acceleration should be recalculated based on the combined forces. The thread highlights the complexity of the problem and the need for careful analysis of the forces involved.
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Homework Statement


The coefficient of kinetic friction between the 2.0 kg block in figure and the table is 0.350.

What is the acceleration of the 2.0 kg block?

Homework Equations


F=uK*Force Normal
Fnet=ma

The Attempt at a Solution


I tried to take the mass's of both the hanging and used them to calculate a net force on the block in the center and then solved for acceleration that way. Here's my work.

3kg*9.8=29.4N Right Side Tension
1kg*9.8=9.8N Left Side Tension
2kg*9.8=19.6N Force Normal

F=.35*19.6N
6.86N

(29.4N-6.86N)-9.8N=2kg*acceleration

a=6.34m/s^2

and it turns out this is wrong please help me out thank you!

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candykane91 said:

Homework Statement


The coefficient of kinetic friction between the 2.0 kg block in figure and the table is 0.350.

What is the acceleration of the 2.0 kg block?


Homework Equations


F=uK*Force Normal
Fnet=ma


The Attempt at a Solution


I tried to take the mass's of both the hanging and used them to calculate a net force on the block in the center and then solved for acceleration that way. Here's my work.

3kg*9.8=29.4N Right Side Tension
1kg*9.8=9.8N Left Side Tension
you are assuming that the tension is equal to the weight. That is true only if the system is not accelearting. This one is accelerating, so you must use free body diagrams of each block and apply Newton's 2nd and 3rd law.
2kg*9.8=19.6N Force Normal

F=.35*19.6N
6.86N
OK
(29.4N-6.86N)-9.8N=2kg*acceleration

a=6.34m/s^2

and it turns out this is wrong please help me out thank you!
isolate each block and solve 3 equations with 3 unknowns.
 
I've got the same question.

I've got these three equations:

m1*a = T1 - m1*g
m2*a = T3 - T1 - μ*m2*g
m3*a = m3*g - T3

Are those right? Solving for a using the first and second equations a=(g(m3-m1))/(m3+m1) which gives a=4.9 which would be the acceleration of the system if the centre mass wasn't there.

Using that value in the net force equation Fnet = m3(g-a) - m1(g+a) - μ*m2*g gives Fnet=6.86. Solving Fnet=ma for a gives you a=3.43 which is not right the right answer.

Can anyone tell me where I went wrong? Thanks!
 
jmm said:
I've got the same question.

I've got these three equations:

m1*a = T1 - m1*g
m2*a = T3 - T1 - μ*m2*g
m3*a = m3*g - T3

Are those right?
Looks OK, although the acceleration is likely to be in the other direction.
Solving for a using the first and second equations a=(g(m3-m1))/(m3+m1)
looks like you canceled out T1 and T3, in error, using equations 1 and 3. T1 and T3 are not equal
which gives a=4.9 which would be the acceleration of the system if the centre mass wasn't there.
This is not true. Correct your method of solving the 3 equations with 3 unknowns.
 
jmm said:
I've got the same question.

I've got these three equations:

m1*a = T1 - m1*g
m2*a = T3 - T1 - μ*m2*g
m3*a = m3*g - T3

Are those right? Solving for a using the first and second equations a=(g(m3-m1))/(m3+m1) which gives a=4.9 which would be the acceleration of the system if the centre mass wasn't there.

Using that value in the net force equation Fnet = m3(g-a) - m1(g+a) - μ*m2*g gives Fnet=6.86. Solving Fnet=ma for a gives you a=3.43 which is not right the right answer.

Can anyone tell me where I went wrong? Thanks!

are you by any chance doing mastering physics?
 
PhanthomJay said:
Looks OK, although the acceleration is likely to be in the other direction. looks like you canceled out T1 and T3, in error, using equations 1 and 3. T1 and T3 are not equal This is not true. Correct your method of solving the 3 equations with 3 unknowns.
Ahh you're a life saver. Thanks a bundle!

candykane91 said:
are you by any chance doing mastering physics?
I sure am. Combine those three equations I posted and you'll get the right answer.
 
jmm said:
Ahh you're a life saver. Thanks a bundle!I sure am. Combine those three equations I posted and you'll get the right answer.

ok. and btw who is this? are you going to u of c? physics 211 or 221?. the anser i got was 32 but i dunt think that's right. I used the 1st and 2nd equations to get T1 and T3 and plugged them into the 2nd equation. can u tell me wat i did wrong?
 
Last edited:
ok now i got 8.2 as in answer. is that right?
 
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