Tension/acceleration between 2 Masses

[tascja]

Homework Statement

two masses are connected by a thing string running over a massless pulley. m1 slides on a 35deg ramp with a coefficient of kinetic friction of 0.40, while m2 hans from the string. what is the acceleration of the masses?
m1 = 1.5 kg
m2 = 3 kg

Homework Equations

Fnet = ma

The Attempt at a Solution

im not quite sure how to relate the object that is on an incline to the object that is hanging and has its forces going up and down?

would the net force equation just be:
Fnet = (m1+m2)a
m2g - Ff - T = (m1 +m2)a

 

[Doc Al]

Rather than attempt to treat both masses together in one equation from the outset, analyze the forces acting on each mass separately. Apply Newton's 2nd law to each, then combine the two equations to solve for the acceleration.

 

[Delphi51]

m2g - Ff - T = (m1 +m2)a

You wouldn't put in the Tension when including all the forces like this. Tension is put in when you use separate equations for each mass.

You do have another force to add to Ff acting on m1 - the component of the force of gravity that is acting down the ramp.

 

[tascja]

Rather than attempt to treat both masses together in one equation from the outset, analyze the forces acting on each mass separately. Apply Newton's 2nd law to each, then combine the two equations to solve for the acceleration.

when you say combine do you mean equate or add the two equations??
Here i equated by isolating T and substituting into the other equation.

so for mass1:
Fnet = ma
ma = T-(downsloping component of gravity) - Ff
= T- 14.7sin35 - 5.88
T = (1.5)a +2.55

for mass2:
Fnet = ma
mg - T = ma
T = mg - ma
T = 29.4 - 3a

then:

29.4 - 3a = (1.5)a + 2.55
a = 5.97 m/s^2

does that seem right?

 

[Doc Al]

when you say combine do you mean equate or add the two equations??

I just meant solve them simultaneously. There are several ways to do that. (Adding them to eliminate T might be the easiest.)

Here i equated by isolating T and substituting into the other equation.

Perfectly fine.

so for mass1:
Fnet = ma
ma = T-(downsloping component of gravity) - Ff

Good.

= T- 14.7sin35 - 5.88
T = (1.5)a +2.55

How did you solve for the friction?

for mass2:
Fnet = ma
mg - T = ma
T = mg - ma
T = 29.4 - 3a

Good.

then:

29.4 - 3a = (1.5)a + 2.55
a = 5.97 m/s^2

does that seem right?

Right idea, but check your numbers.

 

[tascja]

How did you solve for the friction?

** sorry i don't know where i got 5.88 from?? but i think it should be:
in the question it told me that the coefficient of friction is 0.40.
Ff = μFn

Fn = 14.7cos35
= 12.04 N

Ff = (12.04)(0.4)
= 4.816 N

 

[Doc Al]

** sorry i don't know where i got 5.88 from?? but i think it should be:
in the question it told me that the coefficient of friction is 0.40.
Ff = μFn

Fn = 14.7cos35
= 12.04 N

Ff = (12.04)(0.4)
= 4.816 N

Much better.