# Physics Problem: Fnet=ma or Fnet= 0?

• Jaimie
In summary: For instance, if you were to tilt the slope by 1 degree, the box would still not move. If you tilt it by 2 degrees, it still hasn't moved... but when you tilt it by 22 degrees, the box starts to slide.The calculations in the beginning are correct because they help you find the angle at which the box starts to slide (the maximum angle possible without sliding), which is what the question is asking for. The wording just might be a little confusing, but the concept is the same.
Jaimie
This question has been posted and I do apologize for posting again, but I am having trouble determining when to use Fnet=ma or Fnet= 0 for questions like these. I have solved it but need help interpreting the question.

"At a construction site, a small crane is raising two boxes of nails on a plank to the roof. One box has already been opened and is half-full, while the other box is new. The boxes, including the nails, weigh 10kg and 20kg respectively, and are the same size"

"b) If the coefficient of static friction is 0.4, draw and FBD for each box of nails and use it to calculate the angle at which each box begins to slide".

So here, I am getting confused. To me, I am seeing this motion involving acceleration as it is beginning to move. But in order to answer, the problem does not involve Fnet=ma. Why is this?

"c) If the coefficient of kinetic friction is 0.3, how fast will the boxes accelerate along the plank once they start to slide?"

So now, we are involving acceleration, but in b) we are not, and it appears that the language is the same to me (ie. beginning to move vs. starting to slide). It just seems like they are saying the same thing.
Thank you!

Last edited by a moderator:
Well if you imagine the roof as having a variable angle (you can adjust it).

What conditions are necessary for the boxes of nails to NOT move (I'm thinking of one inequality) don't overthink it.

part c is a little confusing. When you change the coeffecient of friction, you change the angle that the nails start to move. Are you supposed to use the angle from B or are you supposed to solve for a new angle in C (which will give you an utterly meaningless acceleration)?

My suggestion to you is to remember that according to Newton's Laws of Force, a body in motion stays in motion and an object at rest stays at rest, unless acted upon by an external force. F = ma, and F = 0 are the mathematical representations of this statement.

F = 0 implies the system's forces are in equilibrium, or that every action has an equal and opposite reaction. This means the system is not being acted on by any outside forces and is either at rest, or moving with CONSTANT velocity (If you have any calculus background you know that a = $\frac{dv}{dt}$, and that the derivative of a constant is 0).

If however the system is not at rest (static) or moving with CONSTANT velocity then there is an outside force present and we we must use F = ma. This equation implies the system is changing and that the once stationary object is now moving, or the moving object is slowing down or speeding up. An acceleration is present in the system causing it to move from equilibrium (where the system wants to be naturally). The "ma" side of this equation is representative of this "outside" force, and the F represents the SUM of the forces present WITHIN the system in order for it to be in equilibrium.

Hope this helps!

Jaimie said:
I am seeing this motion involving acceleration as it is beginning to move. But in order to answer, the problem does not involve Fnet=ma. Why is this?
If it involves acceleration, then why do you say it does not involve Fnet=ma? For there to be acceleration, $F_{net}\neq 0$

BiGyElLoWhAt said:
When you change the coeffecient of friction, you change the angle that the nails start to move. Are you supposed to use the angle from B or are you supposed to solve for a new angle in C (which will give you an utterly meaningless acceleration)?

They didn't change the coefficient of friction, they gave you the kinetic coefficient (while the 1st coeff given was the static coeff)

The "once they start to slide" part of part c implies that you use the angle from part b

Nathanael said:
They didn't change the coefficient of friction, they gave you the kinetic coefficient (while the 1st coeff given was the static coeff)

The "once they start to slide" part of part c implies that you use the angle from part b

Ahhh good catch Nathanael.

@OP
Even when you're using F = 0, you're still using F = ma, the only difference is that a = 0.

To Nathanial and Jaytech (thanks for the clarification!). I solved the problem in the following way:
Fnet= 0
Fgx- Ff= 0
Fgx= sin(theta)(Fg)= sin(theta)(mg)
Ff= Fn(u)= cos(theta)(Fg)(u)= cos(theta)(mg)(u)

sin(theta)(mg)- cos(theta)(mg)(u)= 0
sin(theta)(mg)= cos(theta)(mg)(u) (*mg cancels)
sin(theta)cos(theta)= 0.4
tan(theta)= 0.4
Therefore ..theta= tan-1 (0.4)=21.8= 22 degrees.

So this is the right answer. The words "starting to slide" to me are implying that there is a change in velocity? Am I incorrect? If so, why are the beginning lines to my solution correct?

Jaimie said:
To Nathanial and Jaytech (thanks for the clarification!). I solved the problem in the following way:
Fnet= 0
Fgx- Ff= 0
Fgx= sin(theta)(Fg)= sin(theta)(mg)
Ff= Fn(u)= cos(theta)(Fg)(u)= cos(theta)(mg)(u)

sin(theta)(mg)- cos(theta)(mg)(u)= 0
sin(theta)(mg)= cos(theta)(mg)(u) (*mg cancels)
sin(theta)cos(theta)= 0.4
tan(theta)= 0.4
Therefore ..theta= tan-1 (0.4)=21.8= 22 degrees.

So this is the right answer. The words "starting to slide" to me are implying that there is a change in velocity? Am I incorrect? If so, why are the beginning lines to my solution correct?
At 21.8 degrees, the nails are on the verge of moving, motion is pending, but equilibrium is still being maintained, albeit very unstable equilibrium. At a tiny fraction of a degree above the 21.8 degrees, or with a tiny push, they will slide, and kinetic friction takes over.

Jaimie said:
The words "starting to slide" to me are implying that there is a change in velocity? ... why are the beginning lines to my solution correct?

Setting the force of gravity (down the slope) equal to the (maximum) force of static friction is correct because that gives you the "maximum angle possible without sliding" (or, as you wrote: "Fgx- Ff= 0")

The "maximum angle possible without sliding" is essentially the exact same angle as "the angle where it starts to slide" because if you were to increase the angle by an infinitely small amount, it would start to slide.I'm not sure if that made sense, but feel free to ask more questions.

Jaimie said:
This question has been posted and I do apologize for posting again, but I am having trouble determining when to use Fnet=ma or Fnet= 0 for questions like these. I have solved it but need help interpreting the question.

"At a construction site, a small crane is raising two boxes of nails on a plank to the roof. One box has already been opened and is half-full, while the other box is new. The boxes, including the nails, weigh 10kg and 20kg respectively, and are the same size"

"b) If the coefficient of static friction is 0.4, draw and FBD for each box of nails and use it to calculate the angle at which each box begins to slide".

So here, I am getting confused. To me, I am seeing this motion involving acceleration as it is beginning to move. But in order to answer, the problem does not involve Fnet=ma. Why is this?
Up until the instant static friction releases, the box is in static equilibrium. So the net force on the box is zero. Once static friction releases, the frictional force on the box suddenly drops to a lower value, the box is no longer in static equilibrium, and it begins to accelerate. For each box to begin to slide, static friction must first release.

Chet

jaytech said:
My suggestion to you is to remember that according to Newton's Laws of Force, a body in motion stays in motion and an object at rest stays at rest, unless acted upon by an external force. F = ma, and F = 0 are the mathematical representations of this statement.

Newton didn't formulate Laws of Force; he formulated Laws of Motion.

Thank you Chet and to everyone for your replies. I get it now.

## 1. What is the meaning of Fnet=ma or Fnet=0 in physics?

Fnet=ma is the equation for Newton's second law of motion, which states that the net force (Fnet) acting on an object is equal to the mass (m) of the object multiplied by its acceleration (a). Fnet=0, on the other hand, means that the net force acting on an object is zero, indicating that the object is either at rest or moving with a constant velocity.

## 2. Why is Fnet=ma or Fnet=0 important in physics?

Fnet=ma or Fnet=0 is important in physics because it helps us understand the relationship between force, mass, and acceleration. This equation is fundamental in solving problems related to motion, such as calculating the force required to move an object or determining the acceleration of an object under the influence of a net force.

## 3. How do I apply Fnet=ma or Fnet=0 in solving physics problems?

To apply Fnet=ma or Fnet=0 in solving physics problems, you first need to identify the net force acting on the object and determine its magnitude and direction. Then, you can use the equation Fnet=ma to calculate the acceleration of the object, or use Fnet=0 to determine the force required to keep the object at rest or moving at a constant velocity.

## 4. Can Fnet=ma or Fnet=0 be used for any type of motion?

Yes, Fnet=ma or Fnet=0 can be used for any type of motion, as long as the net force acting on the object is known. This equation is valid for both linear motion (e.g. an object moving in a straight line) and rotational motion (e.g. an object spinning around an axis).

## 5. How is Fnet=ma or Fnet=0 related to the concept of inertia?

Fnet=ma or Fnet=0 is closely related to the concept of inertia, which is an object's tendency to resist changes in its state of motion. Inertia is represented by an object's mass in the equation Fnet=ma, where a larger mass means a greater resistance to acceleration. In the case of Fnet=0, the object's inertia is balanced by the forces acting on it, resulting in either a state of rest or constant velocity.

• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
48
Views
6K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
9
Views
5K
• Introductory Physics Homework Help
Replies
13
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
933
• Introductory Physics Homework Help
Replies
10
Views
4K
• Introductory Physics Homework Help
Replies
7
Views
3K
• Introductory Physics Homework Help
Replies
1
Views
4K