How Do You Calculate the Minimum Force to Prevent Sliding on an Inclined Plane?

In summary, the minimum force required to prevent the crate from sliding down an incline is 256.89 Newtons.
  • #1
cak
15
0
Here is the problem I'm having trouble with...

The coefficient of static friction between the 32 kg crate and the 35.0° incline of Figure 4-34 is 0.2. What is the minimum force, F, that must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?

I did a force diagram and figured out the variables:

1. I got normal Force as 257.152
2. force of friction Fk=µk*N = .2 * 257.152 = 51.43
3. perpendicular force = 32 * 9.8 * cos(35) = 256.89
4. parallel force = 32 * 9.8 * sin(35) = 179.87 = 179.87
5. Net force=parallel force - force friction = 179.87 - 51 = 128.49

But that isn't right. Can someone help me?
 
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  • #2
?? How is your normal Force different from your perpendicular component of force?

You want the force of friction to be equal to the parallel force in order for it not to start sliding, right? How do you increase the force of friction?
 
  • #3
cak said:
1. I got normal Force as 257.152

There is another force F acting (the unknown one) in the direction of the weight. So, what does the normal force equal?
 
  • #4
umm by increasing the normal force or µk??

I think this is right since the force of friction=µk*N
Considering the fact that the force of friction is directly related to µk and/or the normal force.

am i right?

thanx for helping me!
 
  • #5
do you mean like the µk (.2) * normal force has to equal 179.87?
so like (the unknown force + the normal force of 257) * .2 = 179.87?
is that right?
 
  • #6
cak said:
5. Net force=parallel force - force friction = 179.87 - 51 = 128.49

Think about what's wrong in this step. (Hint: if the net force does not equal zero, what does Newton's 2nd law tell you?)
 
  • #7
F=ma ...that if the force goes up then the acceleration goes up?
cause mass stays the same here.
 
  • #8
Newtons second law tells us that acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. acceleration is directly related to the net force and inversely related to the mass...
 
  • #9
cak said:
Newtons second law tells us that acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. acceleration is directly related to the net force and inversely related to the mass...

What is important is that you had a net force calculated, which would mean that you have an acceleration. But you want the crate to rest. So, what does that mean? In what relation do the forces parallel to the incline have to be?
 
  • #10
Since the crate is at rest... doesn't that mean that the acceleration has to be zero?
so would that mean that the forces parallel to the incline would have to be zero...possibly??
 
  • #11
cak said:
Since the crate is at rest... doesn't that mean that the acceleration has to be zero?
so would that mean that the forces parallel to the incline would have to be zero...possibly??

Acceleration is zero. The sum of the two forces equals zero (i.e the net force). Now, what does that mean? (Hint: what does that tell you about their directions and their magnitudes?)
 
  • #12
Wouldnt that mean that the force of friction is equal to the parallel force? Doesnt that also mean that they're going in the opposite direction and their magnitudes are the same...
 
  • #13
cak said:
Wouldnt that mean that the force of friction is equal to the parallel force? Doesnt that also mean that they're going in the opposite direction and their magnitudes are the same...

Yes, exactly. You're on the right track now.
 
  • #14
ok well...
correct me if I am wrong but...
if acceletation is zero, then wouldn't that mean
ma=Fparallel-Ffriction
0=32(-9.8)(sin35)-.2(-9.8)(32)(cos35)

but how do i apply that to get the force applied to prevent the crate from sliding down?
 
  • #15
cak said:
ok well...
correct me if I am wrong but...
if acceletation is zero, then wouldn't that mean
ma=Fparallel-Ffriction
0=32(-9.8)(sin35)-.2(-9.8)(32)(cos35)

but how do i apply that to get the force applied to prevent the crate from sliding down?

Ffriction = mu*N, right? Now, as I already asked before, what does N equal? It's not just 9.8*32*cos35. There is another force involved. Look at the text of your problem.

P.S. You don't have to write the minus sign in front of 9.8
 
  • #16
Does that mean I am adding the unknown force to N?
 
  • #17
Think of N as a reaction force. The forces acting in the direction perpendicular to the incline are the component of the weight of the crate in that direction (m*g*cos(35)), and the unknown force F, in the same direction. Since N is a reaction to these forces, it must balance them and point in the opposite direction. Hence, N = m*g*cos(35) + F. Now you should be able to solve your problem without any further difficulties.
 
  • #18
thank you so much :)
 

Related to How Do You Calculate the Minimum Force to Prevent Sliding on an Inclined Plane?

1. What is an inclined plane?

An inclined plane is a simple machine that is a flat surface set at an angle, with one end higher than the other. It is used to raise or lower objects, making it easier to move them from one height to another.

2. How do you calculate the mechanical advantage of an inclined plane?

The mechanical advantage of an inclined plane is calculated by dividing the length of the incline by its height. This provides the ratio of the force required to lift an object up the incline to the force that would be required to lift it straight up.

3. What is the relationship between the angle of the inclined plane and the effort required to move an object?

The steeper the angle of the inclined plane, the less effort is required to move an object up the incline. However, a steeper angle also means that the distance the object must travel is longer, so the overall work required remains the same.

4. How does friction affect an inclined plane?

Friction is a force that opposes motion, so on an inclined plane, it makes it harder to move an object up or down. The steeper the angle of the incline, the greater the frictional force will be.

5. Can an inclined plane be used to lift an object higher than its starting point?

Yes, an inclined plane can be used to lift an object higher than its starting point. The longer the incline, the higher the object can be lifted. However, the force required to lift the object will also increase.

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