Netwon's second law, friction

In summary, the person in the drawing is standing on crutches and the force exerted on each crutch by the ground is directed along the crutch. The coefficient of static friction between the crutch and the ground is 0.90 and the largest angle θMAX that the crutch can have just before slipping on the floor can be determined using the equations \sum{F_x} = (Something) and \sum{F_y} = (Something). The normal force is both Fcosθ and mg/2, as each crutch supports half of the person's weight. It is important to remember that forces are vectors and have magnitude and direction. If stuck, additional assistance can be provided.
  • #1
jehan4141
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0
The person in the drawing is standing on crutches. Assume that the force exerted on each crutch by the ground is directed along the crutch, as the force vectors in the drawing indicate. If the coefficient of static friction between a crutch and the ground is 0.90, determine the largest angle θMAX that the crutch can have just before it begins to slip on the floor.

nw0173-nu.jpg


μ = 0.90
Ff = (Fn)μ
Untitled.png

I know that I am missing a force for the problem to work, but I have thought about it for awhile and I can't seem to picture it. The normal force is supposed to be a force that is perpendicular to the surface...does that mean the normal force is Fcosθ? Is there supposed to be another force in the 4th quadrant? If yes, what is that force? Is it the force of the man pushing down on the crutch? If yes, does the force equal to F then?
 
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  • #2
jehan4141 said:
The person in the drawing is standing on crutches. Assume that the force exerted on each crutch by the ground is directed along the crutch, as the force vectors in the drawing indicate. If the coefficient of static friction between a crutch and the ground is 0.90, determine the largest angle θMAX that the crutch can have just before it begins to slip on the floor.

View attachment 39317

μ = 0.90
Ff = (Fn)μ


View attachment 39318
I know that I am missing a force for the problem to work, but I have thought about it for awhile and I can't seem to picture it. The normal force is supposed to be a force that is perpendicular to the surface...does that mean the normal force is Fcosθ? Is there supposed to be another force in the 4th quadrant? If yes, what is that force? Is it the force of the man pushing down on the crutch? If yes, does the force equal to F then?

Intersting situation.
Yes Normal force = Fcosθ , but interestingly, the normal force is mg/2 as well, since each crutch has to support half the man's weight.
So as the angle changes, F becomes bigger.
 
  • #3
When dealing with problems like this, it is very important to remember that forces are not scalar numbers, but vectors, and therefore have magnitude and direction. As such you have:

[itex]\vec{F}=F_x \hat{i} + F_y \hat{j}[/itex]

Thus you can have 2 equations:

[itex]\sum{F_x} = (Something)[/itex]
[itex]\sum{F_y} = (Something)[/itex]

Let us know if you get stuck again. We'll be glad to assist. :smile:
 

1. What is Newton's Second Law?

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

2. How does friction affect Newton's Second Law?

Friction is a force that opposes motion and it can affect Newton's Second Law by reducing the acceleration of an object due to the increased force required to overcome friction.

3. What is the formula for Newton's Second Law?

The formula for Newton's Second Law is F = ma, where F is the net force, m is the mass of the object, and a is the acceleration.

4. Can friction ever be beneficial in regards to Newton's Second Law?

Yes, friction can be beneficial in certain situations. For example, friction allows us to walk and drive on surfaces without slipping. It also helps us grip objects and perform tasks such as writing or holding a phone.

5. How does Newton's Second Law apply to real-life scenarios?

Newton's Second Law is applied in many real-life scenarios, such as calculating the force needed to push a car or lift a heavy object. It also helps us understand the motion of objects in sports, such as how much force is needed to kick a soccer ball or hit a baseball.

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