General Forces on Slope Problem

In summary, the equation for m2/m1 = m1(sine(theta) + mu*cos(theta)) simplifies the general expression for mass m2 in terms of (mu) and (theta). To find a numerical value for (mu) and (theta) so that m2/m1 > 1, one can compare small angles to larger angles and make up values that satisfy the equation.
  • #1
candycooke
14
0
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.
 

Attachments

  • General Slope Problem.bmp
    88.6 KB · Views: 599
Physics news on Phys.org
  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
Thank you for your help Doc Al. It all makes sense now.
 

FAQ: General Forces on Slope Problem

What is the concept of "General Forces on Slope Problem"?

The "General Forces on Slope Problem" refers to the analysis of all the forces acting on an object or structure on a slope. This includes the weight of the object, normal force from the slope, and any other external forces such as friction or tension.

How do you calculate the normal force on an object on a slope?

The normal force on an object on a slope can be calculated by taking the component of the force of gravity that is perpendicular to the slope. This can be found using trigonometric functions and the angle of the slope.

What is the significance of "General Forces on Slope Problem" in real-world applications?

Understanding the concept of "General Forces on Slope Problem" is crucial in designing and analyzing structures on slopes, such as roads, buildings, or retaining walls. It helps ensure the stability and safety of these structures and prevents potential hazards.

How does the angle of the slope affect the forces acting on an object?

The angle of the slope directly affects the normal force and friction force on an object. As the angle increases, the normal force decreases, making the object more prone to sliding down the slope. The angle also affects the direction and magnitude of the friction force, which can either assist or resist the object's movement.

Are there any assumptions or limitations when solving "General Forces on Slope Problem"?

One of the main assumptions in solving the "General Forces on Slope Problem" is that the slope is assumed to be uniform and the object is a point mass. This may not accurately represent real-world scenarios, and thus, the solutions may not be entirely accurate. Additionally, other factors such as weather conditions and soil properties may also affect the forces acting on the object.

Similar threads

Back
Top