Solving the Friction of a Crate on a 20° Plane

In summary, the crate is on a 20 degree incline with coefficients of static and kinetic friction of .45 and .35 respectively. The acceleration of the crate can be found by using the equation F - Fk = ma, where F is the net force, Fk is the force of kinetic friction, and a is the acceleration. After calculating, the answer appears to be 0 according to the book, as the net force is less than the static friction force. The static friction force will always oppose the total sum of the forces parallel to the surface it acts from. The only force in the direction of the incline is mg * sin(20), so the force of static friction will try to oppose this. However, it
  • #1
laxboi33
9
0

Homework Statement



A crate is on a 20 degree plane where the coefficients of static and kinetic friction are .45 and .35 respectively. What is the acceleration of the crate?




Homework Equations



Mk= .35
Ms= .45
F= sin20 * mg (I think this is the net force?)



The Attempt at a Solution



F - Fk= M*A

(sin20 * mg) - (.35 * mg) = m * a

a = sin20 * g - .35 * g

a= .34 * 9.8 - .35 * 9.8

a = -.098


The answer in the book says 0. As I was writing this problem I think I figured it out. If sin 20 is the net force, then that would make net force less than the static friction which would give it 0 acceleration. Maybe I just miscalculated the net force. If so could someone explain to me what net force would be for this equation? Thanks, guys.
 
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  • #2


You presupposed that the crate slides down the incline.

You must first check to see if it starts moving at all!

The static friction force is a reaction force. It will always oppose the total sum of the forces parallel to the surface it acts from.

The only force in the direction of the incline, other than friction, is as you've pointed out, [tex]mg\sin{\theta}[/tex]

Now, the force of static friction tries to oppose this. Can it? What is the maximum force that the static friction can oppose? Is the total force it has to oppose less than, or more than this limit?
 
  • #3


thanks cat
 

1. How is friction defined in the context of a crate on a 20° plane?

In this context, friction refers to the resistance force that acts on the crate as it slides down the 20° plane. It is caused by the interaction between the surfaces of the crate and the plane.

2. What factors affect the friction of a crate on a 20° plane?

The factors that affect friction in this situation include the weight of the crate, the angle of the plane, the material and texture of the surfaces, and the presence of any lubricants or other substances on the surfaces.

3. How can the coefficient of friction be determined for a crate on a 20° plane?

The coefficient of friction can be determined by dividing the magnitude of the frictional force by the weight of the crate. This can be measured experimentally or calculated using the known values of the weight, angle, and other factors.

4. How does the angle of the plane affect the friction of a crate?

The friction of a crate on a 20° plane will be greater than on a flat surface due to the increased normal force acting on the crate. As the angle of the plane increases, the frictional force will also increase.

5. What are some strategies for reducing the friction of a crate on a 20° plane?

Some strategies for reducing friction in this situation include using a smoother surface for the plane, adding a lubricant or reducing the weight of the crate. Additionally, adjusting the angle of the plane or using a different material for the crate or plane can also affect the frictional force.

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