Block on an incline and frictional force

In summary, the problem involves a 4.00-kg block on a 30.0 degree incline, with a coefficient of static friction of 0.700. The question asks for the magnitude of a horizontal force needed to start the block moving up the incline. By drawing a force body diagram and using the equations F=ma and Ffriction=μs*N, the correct answer is found to be 84.1 N. The key mistake was drawing the applied force as parallel to the slope, rather than horizontal, which affected the calculation of the normal and frictional forces.
  • #1
Carrie
27
0

Homework Statement


A 4.00-kg block rests on a 30.0 degree incline. If the coefficient of static friction between the block and the incline is 0.700, with what magnitude force must a horizontal force act on the block to start it moving up the incline?

Homework Equations


F=ma

Ffriction=μs*Fnormal

The Attempt at a Solution


I attached the force body diagram I did.

Fy
Fn - mgsin(30) = 0
Fn = mgsin(30)[/B]


Fx

Ffr - FA = 0
Ffr = FA
μ*N = FA
mgsin(30) + .7*mgcos(30)) = FA
(4)(9.8)sin(30)+.7*(4)(9.8)cos(30) = FA
FA = 43.36 N

However, the answer is 84.1 N. I'm pretty rusty with this stuff. Thank you!
Snapshot.jpg
 
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  • #2
Carrie said:
a horizontal force
Consider what does it means to be "horizontal" here.
 
  • #3
Hi Carrie, Welcome to Physics Forums.

Note that the problem stated that the applied force is horizontal. You've drawn it as parallel to the face of the slope. This will make a difference :smile:

Edit: D'oh! sometimes=1 beat me to it!
 
  • #4
Hello and thank you!

I'm sorry, I realize what you mean by me drawing the force diagram incorrectly, but I'm still confused as to what that changes.
 
  • #5
Carrie said:
Hello and thank you!

I'm sorry, I realize what you mean by me drawing the force diagram incorrectly, but I'm still confused as to what that changes.
Some part of the applied force pushes the block upslope, but another component will act to press the block into the slope. What other force will be affected by this?
 
  • #6
The frictional force, right?
 
  • #7
Carrie said:
The frictional force, right?
Yup.
 
  • #8
I thought that's what I represented when I said:

Carrie said:
Ffr - FA = 0
 
  • #9
That's not a valid equation. FA (which is horizontal) and the friction force (which acts along the slope) do not act along the same direction.

The force due to friction is obtained from the normal force and the friction coefficient. The normal force is made up of two contributions: one due to a component of the block's gravitational weight, the other due to a component of the force FA.

So FA increases the total normal force, and hence increases the friction too.
 
  • #10
Thank you!:smile:
 

1. What is a block on an incline and frictional force?

A block on an incline is a physical scenario where a solid object (the block) is placed on a sloped surface. Frictional force is the opposing force that arises due to the contact between the block and the inclined surface.

2. How does the angle of incline affect the frictional force?

The greater the angle of incline, the greater the frictional force. This is because the component of the weight of the block along the inclined surface increases with a steeper angle, resulting in a larger force of friction.

3. How does the mass of the block affect the frictional force?

The mass of the block has no direct effect on the frictional force. However, a heavier block will have a greater weight and therefore a larger normal force, which in turn can increase the frictional force.

4. How does the coefficient of friction affect the frictional force?

The coefficient of friction is a measure of the roughness of the surfaces in contact. The higher the coefficient of friction, the greater the frictional force. This means that rougher surfaces will result in a larger frictional force.

5. How can the frictional force be calculated in a block on an incline scenario?

The frictional force can be calculated using the formula Ff = μN, where μ is the coefficient of friction and N is the normal force. The normal force can be found by resolving the weight of the block into its components parallel and perpendicular to the inclined surface.

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