Why Does Cos(30) Calculate the Correct Frictional Force on a Sliding Crate?

In summary: The parallel component of any force is always the magnitude of the force multiplied by the cosine of the angle between the force and the axis. In this case, the force is the weight, and the angle is 30 degrees. Therefore, the parallel force of friction would be the magnitude of the weight (mass x gravity) multiplied by the coefficient of kinetic friction (0.30) and the cosine of 30 degrees. In summary, to find the frictional force on the crate sliding up the ramp, you would use the equation fk=uk*N, where fk is the frictional force, uk is the coefficient of kinetic friction (given as 0.30), and N is the normal force (equal to the magnitude of the weight times the
  • #1
jfk313
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Homework Statement



Find the frictional force (magnitude and direction) on the crate if the crate is sliding up the ramp.

Frictional force= ?
μk = 0.30
μs = 0.74
mass=10.0 kg
angle of incline 30° to the horizontal

Homework Equations



fk=uk * N



The Attempt at a Solution


I tried doing this problem by drawing t he free body diagram and creating another triangle to find frictional force, as seen in the picture
https://www.youtube.com/watch?v=https://www.youtube.com/watch?v=
. i tried solving for the frictional force by multiplying the kinetic coefficient of friction, gravity and mass with sin(30)... I found out from a friend that cos(30) would give you the correct answer, but I not sure why. So I wanted to understand why the sin(30), giving you the parallel force of friction, is not correct when trying to solve this problem.

 

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  • #2
When you draw out the freebody diagram, try making the x-axis the ramp 30 (tilted axis). Then, break up the weight vector into its x and y components, and you should be able to see that mgsin(\theta)is a force in the same direction as friction, but it is NOT friction. Friction is a separate force on the free body diagram all together.
 
  • #3
jfk313 said:
So I wanted to understand why the sin(30), giving you the parallel force of friction, is not correct when trying to solve this problem.

The mgsin(\theta) term is NOT the parallel force of friction. It is the box's parallel (x-component) of weight.
 

Related to Why Does Cos(30) Calculate the Correct Frictional Force on a Sliding Crate?

1. What is ramp frictional force?

Ramp frictional force is the force that acts between an object and a ramp that is in contact with it. It opposes the motion of the object and is caused by the roughness of the surfaces in contact.

2. How is ramp frictional force calculated?

The ramp frictional force can be calculated by multiplying the coefficient of friction between the two surfaces by the normal force acting between them. This can be expressed as Ff = μ * Fn, where Ff is the frictional force, μ is the coefficient of friction, and Fn is the normal force.

3. What factors affect ramp frictional force?

The factors that affect ramp frictional force include the roughness of the surfaces in contact, the weight of the object, and the angle of the ramp. The coefficient of friction also plays a role, which is dependent on the materials of the surfaces in contact.

4. How does ramp angle affect frictional force?

The angle of the ramp can affect the ramp frictional force in two ways. First, as the angle increases, the normal force acting between the object and the ramp decreases, resulting in a decrease in frictional force. Second, as the angle increases, the force of gravity acting on the object along the ramp also increases, resulting in an increase in frictional force.

5. What is the significance of ramp frictional force?

Ramp frictional force is significant because it plays a crucial role in many real-life scenarios, such as walking on an inclined surface, driving a car up a hill, or pushing a heavy object up a ramp. Understanding and managing ramp frictional force can help us avoid accidents and improve efficiency in various tasks.

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