Multiple Forces Acting on Masses Attached by a String

[javaldesc]

A 2.0 kg block (m1) and a 6.0 kg block (m2) are connected by a massless string. Applied forces, F1 = 12.0 N and F2 = 5.0 N, act on the blocks, as shown in the figure below m1 is on the left side of the system, F1 is pulling on m1 to the left; m2 is to the right, F2 is is pulling m2 to the right (both forces are parallel to the x axis). The coefficient of friction between the blocks and the table is μK = 0.16.
A) Draw a free body diagram for EACH block. [3 points]
B) Write Newton’s Second Law equations for each block. [3 points] C) Determine the acceleration of the blocks. [2 points]
D) What is the tension on the string connecting the blocks? [2 points]

Relevant equations:

ƩF=m*a
F friction=\mu*Fn
F gravity=m*g

Attempt to solve problem:

I am assuming the system is moving in the direction of F1, I believe this is where I made a mistake. Anyway, for m1:
ƩFy=m1*g=Fn1
ƩFx=F1-m1*g*μ-T=m1*a

For m2:
ƩFy=m2*g=Fn2
ƩFx=T-m2*g*μ-F2=m2*a

T=m2*a+m2*g*μ+F2

Replacing T for m1:

m1*a=F1-m1*g*μ-(m2*a+m2*g*μ+F2)
∴
a=(F1-m1*g*μ-m2*g*μ-F2)/(m2+m1)=-0.694 m/s^2

I wasn't provided with a solution to the problem, I am sure I made a mistake. Could use some help

 

[PhanthomJay]

A 2.0 kg block (m1) and a 6.0 kg block (m2) are connected by a massless string. Applied forces, F1 = 12.0 N and F2 = 5.0 N, act on the blocks, as shown in the figure below m1 is on the left side of the system, F1 is pulling on m1 to the left; m2 is to the right, F2 is is pulling m2 to the right (both forces are parallel to the x axis). The coefficient of friction between the blocks and the table is μK = 0.16.
A) Draw a free body diagram for EACH block. [3 points]
B) Write Newton’s Second Law equations for each block. [3 points] C) Determine the acceleration of the blocks. [2 points]
D) What is the tension on the string connecting the blocks? [2 points]

Relevant equations:

ƩF=m*a
F friction=\mu*Fn
F gravity=m*g

Attempt to solve problem:

I am assuming the system is moving in the direction of F1, I believe this is where I made a mistake. Anyway, for m1:
ƩFy=m1*g=Fn1
ƩFx=F1-m1*g*μ-T=m1*a

For m2:
ƩFy=m2*g=Fn2
ƩFx=T-m2*g*μ-F2=m2*a

T=m2*a+m2*g*μ+F2

Replacing T for m1:

m1*a=F1-m1*g*μ-(m2*a+m2*g*μ+F2)
∴
a=(F1-m1*g*μ-m2*g*μ-F2)/(m2+m1)=-0.694 m/s^2

I wasn't provided with a solution to the problem, I am sure I made a mistake. Could use some help

Hi javaldesc, welcome to PF!You picked a strange problem to start off with!
I don't know if there might be an error in the problem or in your description of it, but assuming it is correctly stated, the problem is a 'tricky' one. You get a negative acceleration value no matter which direction you choose, which indicates something else is happening, like you might want to consider whether the system is even moving at all, in which case the friction force would be static and less than its limiting value at impending motion. If you had an answer, we could check out what the book approach was. I think it may have been written incorrectly, based on the questions asked.

 

[javaldesc]

I've attached a screen shot of the problem. It is from last year's Physics Final. I was pretty confused by the problem, it is stated that the coefficient of friction is μk. Therefore, I am definitely sure the system is accelerating. Maybe I described the problem incorrectly, I hope the picture helps! Thank you.

 

[PhanthomJay]

I've attached a screen shot of the problem. It is from last year's Physics Final. I was pretty confused by the problem, it is stated that the coefficient of friction is μk. Therefore, I am definitely sure the system is accelerating. Maybe I described the problem incorrectly, I hope the picture helps! Thank you.

The description was good, but as described, it can not be accelerating. Instead of looking at FBD's of each block, try looking at an FBD of both blocks together, that is, look at a FBD of the entire system as a whole. In this manner, you don't have to consider the tension force in the cable connecting the blocks, since it is internal to the system. Thus, the external forces acting are the applied forces F1 and F2, and the friction forces between the blocks and table. Then using Newton's 2nd Law,
F_{net} = (m_1 + m_2)a
F_{applied} - F_{friction} = (m_1+m_2)a
12 - 5 - F_{friction} = 8a
7 - F_{friction} = 8a

Now if the objects are accelerating, then a must be greater than than 0, and less than 7/8, which means the total friction force must be less than 7. But since uk(N) = .16(80) = about 12 or so, that means the system is at rest. The actual friction force is static and less than us(N).

You should note that the picture shows, by scaling, m1 bigger in size than m2, and also F2 greater than F1, in contradiction to the description. Something went wrong when developing the OP. Everyone, including yours truly, makes errors, but it is a bit disconcerting when presented as a final exam question. They probably tossed it out when grading.

 

[javaldesc]

Can the question be answered then? I mean, even if Net Force is zero, there has to be a tension between the masses right?

 

[PhanthomJay]

Can the question be answered then? I mean, even if Net Force is zero, there has to be a tension between the masses right?

Good question. When you look at the FBD of the system, and knowing a = 0, then the 7 N net applied force must be balanced by a 7N force in the other direction. It must be that all friction forces sum to 7N. But does that mean that the cord has no tension? Draw FBD's of each block and give it some thought.