How Do You Calculate Mass M1 in a Frictionless Incline Problem?

AI Thread Summary
To calculate mass M1 in a frictionless incline problem, the forces acting on both masses M1 and M2 are analyzed using Newton's second law. The equations derived show that M1 can be expressed in terms of M2, gravitational acceleration, and the incline angle. A user initially calculated M1 incorrectly but found the correct answer by applying a different approach suggested by another participant. The discussion emphasizes understanding the coordinate system and acceleration directions for both masses. Ultimately, the collaborative effort led to a clearer understanding of the problem and the correct solution.
NoobeAtPhysics
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Homework Statement



M1 and M2 are two masses connected as shown. The pulley and the incline are frictionless. Find the mass M1, given that M2 (4.00 kg) accelerates downwards at 3.53 m/s2, and that θ = 32.0°.

dYVQnGg.gif


Homework Equations



F = m*a
Fk = u*N

The Attempt at a Solution



Forces on m1

Fy = T - N - m1*g*sin(theta) = m1*a
Fx = N - m1*g*cos(theta) = 0

Forces on m2

Fy = m2*g*T = m2*a


Using equations above,

T = m2(g-a)
N = m1*g*cos(theta)

m2*(g-a) - m1*g*cos(theta)-m1*g*sin(theta) = m1*a

rearranging for m1

m1 =[m2*(g-a)]/[a+g(sin(theta) + cos(theta))]


Plugging in the correct values, I get 1.47kg for my answer which is... wrong :(


Thank you,
Noobe
 
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I get

T - sin(θ)m1 g = m1 a
T - m2 g = -m2 a

Two equation with 2 unknowns. Does that help ?
 
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Thank you Mentz that is very helpful! I get the right answer when I use your equations. Thank you again. =]Question:

T - m2 g = -m2 a

Why is it -m2 a and not just m2 a?
 
NoobeAtPhysics said:
Thank you Mentz that is very helpful! I get the right answer when I use your equations. Thank you again. =]Question:

T - m2 g = -m2 a

Why is it -m2 a and not just m2 a?

Think about Newton's second law: F=ma
What is the acceleration of m1? What is the acceleration of m2?
consider gravity and the coordinate system Mentz114 has chosen when drawing free body diagrams for both blocks. That should answer your question.
 
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Thank you very much guys, I understand it now! :), Physics is fun
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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