# TrigonometryBasic trig problem (finding side)

#### daigo

##### Member
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:

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?

#### Ackbach

##### Indicium Physicus
Staff member
Draw a free-body diagram. What are ALL the forces acting on the wheel? In what directions are they pointed?

#### daigo

##### Member
Okay, I drew the picture a bit more clearer:

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?

#### Ackbach

##### Indicium Physicus
Staff member
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?

#### Jameson

Staff member
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

#### daigo

##### Member
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..

#### Jameson

Staff member
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.

#### daigo

##### Member
Well, I think the only component I was missing was the "normal force," and I believe I had eveything else right, so accordingly:

#### Jameson

Staff member
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.

#### daigo

##### Member
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)