Coefficient of static friction problem

AI Thread Summary
The discussion centers on a physics problem involving a wedge on an inclined plane, where a horizontal force is applied. The user initially struggles with the free body diagram and the equations of motion, particularly the roles of sine and cosine in calculating forces. After some back-and-forth, they clarify the correct equations for normal and frictional forces, leading to a better understanding of the acceleration, which they question as being negative. Additionally, a multiple-choice question about forces in a carnival ride is posed, with the consensus confirming that the normal force acts as the centripetal force while friction assists in keeping riders from falling. Overall, the conversation highlights the importance of accurately applying trigonometric functions in physics problems.
Seung Lee
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Homework Statement


I got this problem:
A wedge of mass m = 36.1 kg is held in place on a fixed plane that is inclined by an angle θ = 21.3° with respect to the horizontal. A force F = 302.3 N in the horizontal direction pushes on the wedge, as shown in Figure 4.25a. The coefficient of kinetic friction between the wedge and the plane is μk = 0.159. Assume that the coefficient of static friction is low enough that the net force will move the wedge.

http://semmedia.mhhe.com/physics/bau...dge/index.html
basically what's there except friction would be the other way around (since its accelerating up the slide.

Homework Equations


FBD
Sum of all forces = ma

The Attempt at a Solution


I did a free body diagram and got:
Summation of forces in y:
Normal force = mg sin theta + Appliedforce sin theta

Then for the forces in x:
mgcos + friction - Aplliedforce sin = 36.1a

I did solve for everything but strangely the acceleration gives me acceleration = 3.2

Can any1 help me if I am doing something wrong? Like getting cos and sin mixed?
 
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Check "Normal force = mg sin theta + Appliedforce sin theta"
mg is in a different direction than F, so how can it have the same coëfficiënt ?
also check the angle. Why do you state 21.3 and the text 22 degrees ?

What is so trange about acceleration 3.2 ?
You sure the picture and your exercise match ? Why do you want to calculate the acceleration if the exercise says the task is to draw an FBD ?

Oh, and: welcome to PF :)
 
Oh! Ok, I think fixing the cos and sin around helped me out
it was mg cost + Forceappliedsin = Normal force (sum of forces in y)
and mg sin + friction - Forceappliedcos = ma

I just have another question. The acceleration here would be negative right? My teacher had told me that itw as positive but I wasn't sure why but I'm also pretty sure it's actually negative.

Thank you for the welcoming too :D

P.S.:
Just a fast MC question I had a doubt with:
The Tornado is a carnival ride that consists of a hollow vertical cylinder that rotates rapidly about its vertical axis. As the Tornado rotates, the riders are pressed against the inside wall of the cylinder by the rotation, and the floor of the cylinder drops away. The force that points upward, preventing the riders from falling downward, is
tension, friction, gravity or normal force.

I'm pretty sure its friction due to the fact that normal would act perpendicular (to the center of the cylinder) and gravity would point downward and well tension would not exist.
Am i right?
 
Seung Lee said:
Oh! Ok, I think fixing the cos and sin around helped me out
it was mg cost + Forceappliedsin = Normal force (sum of forces in y)
and mg sin + friction - Forceappliedcos = ma

I just have another question. The acceleration here would be negative right? My teacher had told me that itw as positive but I wasn't sure why but I'm also pretty sure it's actually negative.

Thank you for the welcoming too :D

P.S.:
Just a fast MC question I had a doubt with:
The Tornado is a carnival ride that consists of a hollow vertical cylinder that rotates rapidly about its vertical axis. As the Tornado rotates, the riders are pressed against the inside wall of the cylinder by the rotation, and the floor of the cylinder drops away. The force that points upward, preventing the riders from falling downward, is
tension, friction, gravity or normal force.

I'm pretty sure its friction due to the fact that normal would act perpendicular (to the center of the cylinder) and gravity would point downward and well tension would not exist.
Am i right?
You are right about your last comment. Note that the normal force is acting as the centripetal force. And the friction force is acting upward.
 
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