Contact forces homework problem

[brad sue]

Hello,

I have a problem and I would like to check my answers:

A mass m₁ rests on top of another mass m₂ , which in turn rests on a frictionless horizontal surface. A light cord is attached to
m₂ ,which is used to pull on it witha force F.

a) Find the acceleration ofeach object when the surface between the two objects is frictionless.
I found for m₁ : a₁ =0 ( since frictionless between the surface of the 2 objects)
For m₂ : a₂ =F/ m₂

b) Find the acceleration of each object when the surface the two objects is rough enough to ensure that m₁ does not slide on m₂

I found for m₁ : a₁ = -f₁ / m₁ where f₁ is the friction force due to the contact of the 2 masses.

For m₂ = ( f₁ + F) / m₂

c) What are the magnitude and direction of the contact forces exerced by the lower object on the upper one assuming that the upper object is sliding on the lower object with a non zero coefficient of kinectic μ_k?

For this question isn't the same as the precedent question. we just replace
f₁= μ_k * N?
Thank you for your time

B

 

[mezarashi]

b) Find the acceleration of each object when the surface the two objects is rough enough to ensure that m₁ does not slide on m₂

I found for m₁ : a₁ = -f₁ / m₁ where f₁ is the friction force due to the contact of the 2 masses.

For m₂ = ( f₁ + F) / m₂

I think you have a typo there for the m2 part. Anyway, how I would solve it is:

F = (m1 + m2)a

Since the objects are one and whole, their accelerations are the same. No need to do separate calculations

c) What are the magnitude and direction of the contact forces exerced by the lower object on the upper one assuming that the upper object is sliding on the lower object with a non zero coefficient of kinectic μ_k?

For this question isn't the same as the precedent question. we just replace
f₁= μ_k * N?
Thank you for your time

B

This one doesn't even require the coefficient. The mass $$m_2$$ is on $$m_1$$, so undeniably, the force that is giving $$m_2$$ the acceleration forward is the frictional force. Thus, knowing it's acceleration from part b.

Ffric = Fm2 = m2 x a = μFn = μgm2

^_^

 

[brad sue]

I think you have a typo there for the m2 part. Anyway, how I would solve it is:

F = (m1 + m2)a

Since the objects are one and whole, their accelerations are the same. No need to do separate calculations



This one doesn't even require the coefficient. The mass $$m_2$$ is on $$m_1$$, so undeniably, the force that is giving $$m_2$$ the acceleration forward is the frictional force. Thus, knowing it's acceleration from part b.

Ffric = Fm2 = m2 x a = μFn = μgm2

^_^

Thanks a lot mezarashi

 

[Doc Al]

b) Find the acceleration of each object when the surface the two objects is rough enough to ensure that m₁ does not slide on m₂

I found for m₁ : a₁ = -f₁ / m₁ where f₁ is the friction force due to the contact of the 2 masses.

For m₂ = ( f₁ + F) / m₂

You have the wrong sign for the friction force. The friction on m1 is in the same direction as the acceleration:
$$a_1 = f/m_1$$
$$a_2 = (F - f)/m_2$$
Of course, $a_1 = a_2 = a$, since the two blocks move together. You can solve for the acceleration.

Much easier is to realize that $a = F/(m_1 + m_2)$, as mezarashi suggested.