Physics: Post your doubts here!

[Hamnah Zahoor]

Whenever an object is moving downward we take the weight of the object to be greater than the tension in the string, and when the weight is moving upward the tension is greater than the weight we use this concept while writing the equations.
Yes, T is equal in both cases for both objects are attached to the same string and the string used is usually a light inextensible string which exerts same forces on both masses attached.

 

[Clark20]

Whenever an object is moving downward we take the weight of the object to be greater than the tension in the string, and when the weight is moving upward the tension is greater than the weight we use this concept while writing the equations.
Yes, T is equal in both cases for both objects are attached to the same string and the string used is usually a light inextensible string which exerts same forces on both masses attached.

Isnt mass M pulling mass m and hence exerting the tension force on it? Or is it just the string which is exerting forces of tension on both the masses?

Like, are both the masses completely independent of each other?

 

[Hamnah Zahoor]

Isnt mass M pulling mass m and hence exerting the tension force on it? Or is it just the string which is exerting forces of tension on both the masses?

Like, are both the masses completely independent of each other?

Yes mass ' m ' is pulling mass M which is exerted in the form of tension.
the mass which is heavier exerts the force on the other when it is released.

 

[Clark20]

Yes mass ' m ' is pulling mass M which is exerted in the form of tension.
the mass which is heavier exerts the force on the other when it is released.

Then how is tension exerted on the heavier mass?

 

[Hamnah Zahoor]

Then how is tension exerted on the heavier mass?

At the start you had both masses stationary then you released the system i.e whatever support was holding both masses was removed resulting in heavier mass going down with lighter mass going upwards simultaneously.

 

[Clark20]

At the start you had both masses stationary then you released the system i.e whatever support was holding both masses was removed resulting in heavier mass going down with lighter mass going upwards simultaneously.

Yes, I get that. I understood that the heavier mass is exerting force of tension on the light mass, so it's known where tension acting on the lighter mass comes from. What I dont understand is why is there tension acting on the heavier mass? What exerts it on the heavier mass?

 

[Hamnah Zahoor]

Yes, I get that. I understood that the heavier mass is exerting force of tension on the light mass, so it's known where tension acting on the lighter mass comes from. What I dont understand is why is there tension acting on the heavier mass? What exerts it on the heavier mass?

Why are you making the question so complicated?

I don't know if I am right here
The string due to inertia will resist this change in motion and will impose an opposite force to stop the it.
Thus, the tension in the string acting on the heavier mass.

 

[Clark20]

Why are you making the question so complicated?

I don't know if I am right here
The string due to inertia will resist this change in motion and will impose an opposite force to stop the it.
Thus, the tension in the string acting on the heavier mass.

I'm just trying to be able to write the heavier mass' equation (Mg-T=Ma) with proper understanding of why even there exists force of tension on it.

From my knowledge of force of tension, it is an upward force exerted by a string/rope on an object when its pulled upwards by it. Hence, it made more sense why the lighter mass experiences force of tension as it actually is being pulled upwards by the string connected to the heavier mass. Heavier mass isnt being shown as being pulled by the string, but the lighter mass opposes it from going down freely and probably that way it's exerting force of tension on it.

 

[Hamnah Zahoor]

From my knowledge of force of tension, it is an upward force exerted by a string/rope on an object when its pulled upwards by it. Hence, it made more sense why the lighter mass experiences force of tension as it actually is being pulled upwards by the string connected to the heavier mass. Heavier mass isnt being shown as being pulled by the string, but the lighter mass opposes it from going down freely and probably that way it's exerting force of tension on it.

Agreed

 

[Clark20]

Need help with this question.

 

[Hamnah Zahoor]

Need help with this question.

T - m1g = m1(a)
T = m1(a) + m1g -------(1)

m2g - T = m2(a)
T = m2g -m2(a) --------(2)

placing both equations equal to each other
m1a + m1g = m2g - m2a
m1a + m2a = m2g - m1g
(m1 + m2 )a= (m2-m1) g
a = (m2-m1)g/(m1+m2)

As the question asks about the rate provided by the energy thus,
P = Fv
P = (ma) v

As in the whole system the elevator is going upwards against gravity thus acceleration is negative
P = m(-a)v
P = (m1 + m2)(-(m2-m1)g/m1+m2)v
cancelling out the common terms
P = -(m2-m1)gv
P = (m1-m2)gv ---------- option D

 

[Hamnah Zahoor]

Need help with this question.

If you want a different approach to the answer then visit the following website:

http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-90.html

 

[Clark20]

T - m1g = m1(a)
T = m1(a) + m1g -------(1)

m2g - T = m2(a)
T = m2g -m2(a) --------(2)

placing both equations equal to each other
m1a + m1g = m2g - m2a
m1a + m2a = m2g - m1g
(m1 + m2 )a= (m2-m1) g
a = (m2-m1)g/(m1+m2)

As the question asks about the rate provided by the energy thus,
P = Fv
P = (ma) v

As in the whole system the elevator is going upwards against gravity thus acceleration is negative
P = m(-a)v
P = (m1 + m2)(-(m2-m1)g/m1+m2)v
cancelling out the common terms
P = -(m2-m1)gv
P = (m1-m2)gv ---------- option D

What power are we asked to calculate in the question?

Also, can you please explain how you used the formula P=fv? Why did you substitute those values?

 

[Hamnah Zahoor]

What power are we asked to calculate in the question?

Power = energy/time
Power is the rate at which energy is transferred
thus, we use the power formula here.

 

[Clark20]

Power = energy/time
Power is the rate at which energy is transferred
thus, we use the power formula here.

We are told to calculate the power of the system/masses.
The formula P=FV is used. What do you substitute for m in (ma×v)?

 

[Hamnah Zahoor]

We are told to calculate the power of the system/masses.
The formula P=FV is used. What do you substitute for m in (ma×v)?

(m1 + m2) ------- total mass

 

[Clark20]

(m1 + m2) ------- total mass

Thank you.

 

[Hamnah Zahoor]

Thank you.

You are welcome

 

[Clark20]

T - m1g = m1(a)

As in the whole system the elevator is going upwards against gravity thus acceleration is negative
P = m(-a)v
P = (m1 + m2)(-(m2-m1)g/m1+m2)v
P = (m1-m2)gv ---------- option D

Sorry, but can you please explain this bit again. Why are we considering the direction of acceleration of just the elevator when we are calculating the power of not just the elevator, but the combined power of both the masses/power of the system.

 

[Hamnah Zahoor]

Sorry, but can you please explain this bit again. Why are we considering the direction of acceleration of just the elevator when we are calculating the power of not just the elevator, but the combined power of both the masses/power of the system.

Oh!!! .......yes you are right I made a mistake there.