Atwood machine with friction on horizontal mass

[Mightiestmike]

Ok, if you had a question where a cart (m1) is resting on a surface, and is attached to a string passing over a frictionless pulley, off of which another mass (m2) is suspended in the air, what is the equation that I use?

It is like an atwood machine, except m1 is on a surface, and, I need to take into account the friction affecting m1.

it is hard cause my tect book only shows me how to derive an equation from
Fnet = ma when there is no friction on the surface. But I know it will have friction in the exam.

 

[quasar987]

Ok, if you had a question where a cart (m1) is resting on a surface, and is attached to a string passing over a frictionless pulley, off of which another mass (m2) is suspended in the air, what is the equation that I use?

You use the one and only equation governing all of mechanics. $\Sigma \vec{F} = m\vec{a}$.

It is like an atwood machine, except m1 is on a surface, and, I need to take into account the friction affecting m1.

it is hard cause my tect book only shows me how to derive an equation from
Fnet = ma when there is no friction on the surface. But I know it will have friction in the exam.

Is your problem that you don't know what the form of the frictional force is?

Post your work if you need assistance. And also post exactly what the question is; i.e. what do they want you to find and what do you know about the initial conditions.

 

[Mightiestmike]

Solved it! I think.

For the Mass on the Surface:

a = (T - Ff) / m1

For the Mass suspended in the air:

a = (Fg - T) / m2

I am not sure which is correct, is a = (Fg - T) / m2 correct and you don't take m1 into account because the force on that is caused by T which is already in the equation? or must you put it in the equation like that of an Atwood Machine?

a = (Fg - T) / (m1 + m2)


Atwood Machine:
a = (m2-m1)g / (m1 + m2)

I have a Feeling that the First 2 are correct because I found them in my notes, however I could have copied them incorrectly.

 

[quasar987]

Yes, these are the correct equations of motion. You may combine them to find the tension T, knowing from experience that the acceleration a, is the same for both block.