General Forces on Slope Problem

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Homework Help Overview

The discussion revolves around a physics problem involving forces acting on a mass on a slope, specifically focusing on finding a general expression for the mass ratio of two objects, m2 and m1, under certain conditions of friction (mu) and angle (theta). The original poster seeks clarification on how to assign numerical values to mu and theta to satisfy specific inequalities regarding the mass ratio.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the masses and the forces acting on them, questioning how to derive numerical values for mu and theta without knowing the specific masses. There is a focus on the implications of the equations derived and how they relate to the conditions of the problem.

Discussion Status

Several participants have contributed to refining the equations and clarifying the role of friction and angle in determining the mass ratio. There is an ongoing exploration of how to manipulate the variables to achieve the desired conditions, with hints provided for testing different angle values.

Contextual Notes

Participants note the challenge of assigning numerical values to mu and theta while both masses remain part of the equations. The discussion reflects the constraints of the problem setup and the need for further clarification on the assumptions involved.

candycooke
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a) Find a general expression for mass m2 so that m1 will move at a constant speed up the ramp. Your answer will be based on (mu) and (theta).
b) Give a numerical value for (mu) and (theta) so that m2/m1 > 1
c) Give a numerical value for (mu) and (theta) so that m2/m1 < 1


Fg2 = m2g
Fg1x = m1gsin(theta)
Ff = (mu)Fn
Fn = Fg2y
a = 0m/s2
Fnet = ma
g = 9.8 m/s2

Fnet = Fg2 - Fg1x - Ff
0 = m2g - m1gsin(theta) - (mu)cos(theta)
m2g = m1gsin(theta) + (mu)cos(theta)
m2 = [m1gsin(theta) + (mu)cos(theta)] / g

My problem is that I don't understand how to give a numerical value for (mu) and (theta) so that m2/m1 > or < 1 without also determining the mass of either m1 or m2 since they are both a part of the general equation. Is it even possible without also giving a numerical value to one of the masses?? Any clarification is greatly appreciated.
 

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candycooke said:
0 = m2g - m1gsin(theta) - (mu)cos(theta)
That last term--the friction force--is missing a few factors. Redo it and the solution may be clearer.
 
0 = m2g - m1gsin(theta) - (mu)m1gcos(theta)
m2g = m1gsin(theta) + (mu)m1gcos(theta)
m2 = [m1gsin(theta) + (mu)m1gcos(theta)] / g

Although the equation makes a bit more sense, I'm still confused about how to give a numerical value for (mu) and (theta) so that m2/m1 > or < 1 without also determining the mass of either m1 or m2 since they are both a part of the equation.
 
candycooke said:
0 = m2g - m1gsin(theta) - (mu)m1gcos(theta)
m2g = m1gsin(theta) + (mu)m1gcos(theta)
m2 = [m1gsin(theta) + (mu)m1gcos(theta)] / g
Good. Now further simplify that last expression: Start by factoring out the m1 and the g.
 
m2 = [m1gsin(theta) + (mu)m1gcos(theta)] / g
m2 = [m1g[sin(theta) + (mu)cos(theta)]]/g
m2 = m1[sin(theta) + (mu)cos(theta)]

So factoring has eliminated g but m1 and m2 are sill part of the equation, leaving me just as confused as before.
 
You're almost there. Write it as m2/m1 = ?

Now you get to play around with theta and mu. Hint: Compare a small angle (~ 5 degrees) with a bigger angle (~ 85 degrees).

Remember: All you need to do is make up a few values that satisfy the requirements.
 
Thank you for your help Doc Al. It all makes sense now.
 

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