Basic trig problem (finding side)

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SUMMARY

The discussion revolves around calculating the force required to prevent a 600 lb. wheel from rolling down a 30-degree inclined ramp. The correct approach involves understanding the forces acting on the wheel, specifically the gravitational force, normal force, and the parallel component of gravity acting down the incline. The participant initially misapplied the sine function and neglected to account for the normal force, which is perpendicular to the ramp. The correct calculation yields a force of 300 lbs. to keep the wheel stationary.

PREREQUISITES
  • Understanding of basic trigonometric functions, particularly sine and cosine.
  • Knowledge of free-body diagrams and how to represent forces acting on an object.
  • Familiarity with the concept of normal force in physics.
  • Basic principles of gravity and its effects on objects on inclined planes.
NEXT STEPS
  • Study the concept of normal force and its role in inclined plane problems.
  • Learn how to draw and interpret free-body diagrams for various physical scenarios.
  • Explore the relationship between mass, weight, and gravitational force, including unit conversions.
  • Practice solving inclined plane problems using different angles and weights to reinforce understanding.
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Students studying physics, particularly those tackling problems involving inclined planes, as well as educators seeking to enhance their teaching methods for these concepts.

hatelove
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A 600 lb. wheel is set on a ramp inclined 30 degrees. What is the force required to keep the wheel from rolling down the ramp?

I set the problem like this:

UhteU.png


And I thought the side labeled 'x' is the force needed to keep the wheel up there, which, if calculated by sin(3) = 600/x is 1,200 lbs. but that's not the correct solution.

What am I doing wrong here?
 
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Draw a free-body diagram. What are ALL the forces acting on the wheel? In what directions are they pointed?
 
Okay, I drew the picture a bit more clearer:

XPNHj.png


Gravity is pulling the object straight downwards, and the ramp is affecting the gravity's pull which makes the weight roll down the ramp, so the two forces are gravity pulling the object down, and the object itself going down the ramp. But this is essentially the same as how I depicted it initially, isn't it?
 
There are three forces on the wheel: gravity, the normal force of the ramp on the wheel, and the force you are required to find: pushing up the ramp. Since you want the wheel not to be moving, all the forces must sum to zero. How does this look?
 
This is tricky if you've never seen the formulas before. Even when you've seen the formulas it's easy to mix which trig function to use if you don't focus. You are correct that there is a force of gravity point directly down from the wheel towards the ground and part of this force is directed down the plane, usually called the parallel component. The other component from the force of gravity doesn't go up or down the inclined plane but it involves the wheel and the plane. Any ideas? Whenever you stand on a surface you have the same kind of situation between you and the floor or ground.

Apologies to Ackbach if this is giving away more than he intended but this page I found really nicely shows the whole scenario and how that leads to the various formulas.

http://zonalandeducation.com/mstm/physics/mechanics/forces/inclinedPlane/inclinedPlane.html
 
I had to view a Physics video to understand what normal force was. I didn't know such a thing existed, so now I know there is a force perpendicular to the ramp from the wheel...knowing this I was able to create a smaller triangle within the ramp/wheel and got the correct answer. I don't think it would've made any sense no matter how long I studied this if I hadn't learned basic Physics, so I don't even know why I'm given such problems without being told I need to study the laws of Physics..
 
I think it would be nice to see your solution so others can use this thread in the future. The free body diagram for an inclined plane is not something I would expect a beginning physics student to draw correctly without any help. It appears that your instructor is not covering all necessary background material before assigning problems, which is very frustrating to the student.
 
Well, I think the only component I was missing was the "normal force," and I believe I had eveything else right, so accordingly:

oLyoN.png
 
I believe you are correct that the force you need to consider in order to stop the wheel from rolling is the parallel force down the incline, which uses sin(theta) as you also stated. It looks like you did forget to multiply by gravity though. That should yield the correct answer as far as I can see and be sure to indicate that the force must act the opposite direction that the wheel will go if allowed to roll.

I'm glad for you that you don't have to do any problems with inclines and friction. These problems can get really tedious, or at least they were for me when taking physics in high school and college.
 
  • #10
Is gravity the 9.81 m/s2 formula? We weren't introduced to this because this is a Algebra/Trignometry class, but for a proper Physics solution I guess the gravity would have to be considered (the solution given was 300 lbs. of force so I guess we didn't have to)
 
  • #11
Where I'm from I believe the pound is defined as 2.205 pounds = 1 kilogram, making it a unit of mass. The pound-mass and pound-force units still are strange to me so I just prefer problems in the metric system. If in pounds, then the acceleration of gravity is about 32.17 ft/s^2 and you would need to multiply by this to get units of force. It sounds like it doesn't matter though for this class so I won't persist.
 
  • #12
Note: if you are given that the force of gravity is ____ pounds, you do not have to consider the 32. The pounds figure is equal already to "mg", in British units.
 

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