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

[MrBeans]

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

 

[jbriggs444]

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.

F = mg = 2(10) = 20N

So here, F is the force of gravity on the dangling 2 kg mass.

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.

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.

 

[hutchphd]

Because you did not draw a free body diagram for each mass you have neglected a force. Do it correctly please.

 

[MrBeans]

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.