A 4kg mass sits on a frictionless table....

AI Thread Summary
The discussion revolves around calculating the tension in a string connected to a 4 kg mass on a frictionless table, influenced by a 20 N force and a 2 kg mass hanging off the side. The initial calculations presented by the user were incorrect due to misapplication of formulas and order of operations. The tension was calculated as 13.4 N, but the correct approach requires considering both masses and their respective forces. Clarifications emphasized the importance of drawing free body diagrams and avoiding variable name confusion. Ultimately, the user acknowledged the mistake and expressed gratitude for the guidance.
MrBeans
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Homework Statement
A 4kg mass sits on a frictionless table. A string is attached and leads to the right off the table, over a pulley where it then supports a 2 kg mass. The 4 kg mass is being pulled to the left along the table by a 20 N force. What is the tension in the connecting string?
A. 13N
B. 19.7N
C. 6.4N
D. 35N
Relevant Equations
F = ma
T = mg - ma
I got 13N but is that right because apparently, it's wrong
Here's my work:
F = mg = 2(10) = 20N
F = ma
a = F/m = 20/4+2 = 20/6 = 10/3 = 3.3m/s^2
T = mg - ma T = (2kg)(10m/s^2) - (2kg)(3.3m/s^2) = 13.4 N

I appreciate it! And if I'm wrong could you show how you got your answer? Thanks
 
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MrBeans said:
Homework Statement:: A 4kg mass sits on a frictionless table. A string is attached and leads to the right off the table, over a pulley where it then supports a 2 kg mass. The 4 kg mass is being pulled to the left along the table by a 20 N force.
Let me do this quickly in my head. Indeed, I get something different. So let us check your work.

As I understand it, we have two forces on the 4 kg mass. One of 20 N is applied directly with no mechanism specified. The other from the tension in the string.

MrBeans said:
F = mg = 2(10) = 20N
So here, F is the force of gravity on the dangling 2 kg mass.

MrBeans said:
F = ma
But here, F is the net force on... something.

It is wise to avoid using the same variable name in the same problem to denote two different quantities.

MrBeans said:
a = F/m = 20/4+2 = 20/6 = 10/3 = 3.3m/s^2
Apparently that something has a mass of 4+2. So it is the sum of the two masses. But you only counted one of the two forces.

Note that you wrote 20/4+2 where you presumably meant 20/(4+2) = ##\frac{20}{4+2}##. Order of operations is important.
 
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Because you did not draw a free body diagram for each mass you have neglected a force. Do it correctly please.
 
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jbriggs444 said:
Let me do this quickly in my head. Indeed, I get something different. So let us check your work.

As I understand it, we have two forces on the 4 kg mass. One of 20 N is applied directly with no mechanism specified. The other from the tension in the string.So here, F is the force of gravity on the dangling 2 kg mass.But here, F is the net force on... something.

It is wise to avoid using the same variable name in the same problem to denote two different quantities.Apparently that something has a mass of 4+2. So it is the sum of the two masses. But you only counted one of the two forces.

Note that you wrote 20/4+2 where you presumably meant 20/(4+2) = ##\frac{20}{4+2}##. Order of operations is important.
oh, I got it. Thank you thank you.
 
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