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

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Homework Help Overview

The problem involves two masses, M1 and M2, connected over a frictionless incline and pulley system. The objective is to determine the mass M1 given that mass M2 (4.00 kg) accelerates downwards at 3.53 m/s², with an incline angle of θ = 32.0°.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the forces acting on both masses and derive equations based on Newton's laws. Questions arise regarding the signs in the equations and the implications of the coordinate system used in the free body diagrams.

Discussion Status

Some participants have found helpful guidance in the equations provided by others, leading to a better understanding of the problem. There is an ongoing exploration of the reasoning behind the signs in the equations, indicating a productive dialogue.

Contextual Notes

Participants are considering the implications of the coordinate system and the acceleration of the masses in relation to the forces acting on them. There is an acknowledgment of the need for clarity in the setup of the problem.

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 ?
 
Last edited:
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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
 

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