Solving Neuton's Second Law: Mass and Tension Calculations for Hanging Objects

[chocolatelover]

Homework Statement

Two objects of unequal mass are hung vertically over a frictionless pulley of negigible mass. Determine vector a1, vector a2, and the tentions. m1<m2

Homework Equations

ΣFy=t-m1g=m1ay
ΣFy=m2g-T=m2ay

-m1g+m2g=m1ay+m2ay

ay=m2-m1/(m1+m2)g

The Attempt at a Solution

ay=(m2-m1)/(m1+m2)h

ay=(15-5kg)/(15kg+5kg)(-9.8m/s) (Would gravity be negative?)

=4.9m/s^2

Is a1= a2? that is, a1=-4.9m/s^2 and a2=-4.9m/s^2

T=m1(g+ay)=(2m1m2)/(m1+m2)g

T=5kg(9.8m/s^2+9.8m/s^2+4.9m/s^2)
=2(5kg)(15kg)(9.8m/s^2)/(5kg+15kg)

T=73.5kg

Does this look right?

Would you also get the same answer if you did vector a =change in velocity/change in time?

Thank you very much

 

[kdv]

Homework Statement

Two objects of unequal mass are hung vertically over a frictionless pulley of negigible mass. Determine vector a1, vector a2, and the tentions. m1<m2

Homework Equations

ΣFy=t-m1g=m1ay
ΣFy=m2g-T=m2ay

You should make really clear what axis you are using. Is it clear to you what are the directions you are using? It's a bit confusing because you use two opposite directions for the two objects. But you can do that as long as everything is clear to you.

-m1g+m2g=m1ay+m2ay

ay=m2-m1/(m1+m2)g

The Attempt at a Solution

ay=(m2-m1)/(m1+m2)h

ay=(15-5kg)/(15kg+5kg)(-9.8m/s) (Would gravity be negative?)

=4.9m/s^2

well, you made two mistakes which by luck canceled out, leaving you with the correct answer. You should have used +9.8 for g. And you made a mistake in going from the first to the second line since you dropped the minus sign.

Is a1= a2? that is, a1=-4.9m/s^2 and a2=-4.9m/s^2

Are you talking about $$(a__y)_1$$ and $$(a_y)_2$$? To answer your question, you must make clear what your directions are for your two axis.
And why do you now have a minus sign when there was no minus sign on the previous line?

T=m1(g+ay)=(2m1m2)/(m1+m2)g

T=5kg(9.8m/s^2+9.8m/s^2+4.9m/s^2)
=2(5kg)(15kg)(9.8m/s^2)/(5kg+15kg)

T=73.5kg

Does this look right?

Would you also get the same answer if you did vector a =change in velocity/change in time?

Thank you very much

A tension is a force so it's not in kg.

 

[chocolatelover]

Thank you very much

It's along the y-axis.

I ended up getting kg(kg)(m/s^2)/(kg). Would that be kgm/s^2?

Is ay the same thing as vector a1? Vector a1 and a2 are equal, right? even though m1<m2, would the tention and gravity be the same?

Does it look correct now?

Thank you

 

[kdv]

Thank you very much

It's along the y-axis.

I ended up getting kg(kg)(m/s^2)/(kg). Would that be kgm/s^2?

Yes, and that's the same as a Newton.

Is ay the same thing as vector a1? Vector a1 and a2 are equal, right? even though m1<m2, would the tention and gravity be the same?

There are three things involved and people keep mixing them up which causes a lot of confusion. You may talk about the vectors $$\vec{a}_1, \vec{a}_2$$, you may talk about their components $$(a_y)_1, (a_y)_2$$ and you may talk about their magnitudes (usually written as $$a_1, a_2$$).

The two acceleration vectors are clearly not equal! But the two magnitudes are equal. As for the y components, it depends on the choice of y-axis for each object.

Does it look correct now?

Thank you

It's not clear because you jump from positive to negative to positive values of the y acceelration.

 

[chocolatelover]

Thank you very much

Regards