The maximal normal force (Friction with slanted surface)

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  • #1
Crunge
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Homework Statement:
A box with the mass m is carried by a truck. The truck is driving in a slanted curve with the speed v. The radius of the curve is ρ and its slope is at the angle θ. There is friction between the box and the truck, and the box is not moving. Calculate using the gravitational acceleration g = 9,806.

What is the maximal normal force if the static friction coefficient (μs) is 0,434, m is 5,8 kg and θ is 9,9°?
Relevant Equations:
Maximal friction force (Fmax) = μs * |N|
Truck-problem.jpg


The problem that I immediately ran into was how I would calculate N without knowing Fmax. I didn't think the y-component of N would simply be the same magnitude as mg. After being stuck for a good while I even tested if it was, by dividing the magnitude of mg with cosθ, which of course ended up being wrong. The y-component of Fmax would have to be taken into consideration as well, but to calculate Fmax I wanted to use the formula Fmax = μs * |N|, and since N is also unknown I was stumped again. I then realized the existence of the force P. After snooping around in our study material I found a formula that seemed relevant to this situation and which was similar to Fmax = μs * |N|. According to this, P = μs * mg. However, even after calculating P I'm still not sure what conclusions I can draw about F or N. At first I just tried assuming that F is equal to P but in the opposite direction, however this seemed a bit pointless since then they would take out each other in regards to equilibrium and not affect N right? I decided to ignore P and try my original plan of adding the y-component of F (now equal to P) to mg, so (F * sin9,9 + m*g) and letting this be the y-component of N. Divide that by cos9,9 and I get real close (If rounded down to the proper amount of significant figures it's even correct) but I could see that I had gone about it wrong. I then noticed that in the example for P = μs * mg, P and mg are perpendicular. So I let the old value for P merely be the x-component. Divide that by cos9,9 and I get a new slightly larger value for P. This brought me even closer to the right answer but as espected it is still wrong. After all I'm pretending that P has somehow vanished after F has been calculated. I don't think I can simply let F and P be of equal magnintude, but as long as both F and N are unknown I don't know where to get anymore crucial information out of the problem. This is where I am currently stuck.
 

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  • #2
BvU
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Hi,

Long story, hard to distinguish essentials ...

Question:

What is the trajectory that the box describes ?

So what's the role played by ##\rho## ?

And: which way is ## N## pointing ?

[edit] you might also need some more equations; any idea which ones ?

##\ ##
 
  • #3
Crunge
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Hi,

Long story, hard to distinguish essentials ...

Question:

What is the trajectory that the box describes ?

So what's the role played by ##\rho## ?

And: which way is ## N## pointing ?

[edit] you might also need some more equations; any idea which ones ?

##\ ##
So the problem describes a truck carrying a box taking a curve (which is also slanted), It takes the curve at such speed that the box would start sliding up to the right in the picture if it were not for the friction force F. So P is a centripetal force that F counteracts. (I hope that it's correct to call P a centripetal force, but in other words the truck is taking a curve and I guess since the box has inertia it would slide to the right and that "force" is P). The problem does not mention any trajectory for the box but I believe it should be on the verge of acceleration and is thus stationary. (Stationary as far as this problem is concerned, of course it is moving along the truck in the curve.) N is pointing perpendicularly to the ground/the truck, since the curve is slanted with angle 9,9°. I hope this was what you were wondering about.
 
  • #4
jack action
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The force ##N## drawn on your figure is wrong. By definition, ##N## is the force perpendicular to the friction force.

The force ##P## you put on your Free Body Diagram (FBD) seems to come from nowhere ... and it is.

The truck is taking a curve, therefore there must be an acceleration in your FBD ... you don't have one.
 
  • #5
Crunge
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The force ##N## drawn on your figure is wrong. By definition, ##N## is the force perpendicular to the friction force.

The force ##P## you put on your Free Body Diagram (FBD) seems to come from nowhere ... and it is.

The truck is taking a curve, therefore there must be an acceleration in your FBD ... you don't have one.
Ok so, I don't really know how I should go about depicting P. Since the truck is taking a curve, the box is supposed to be sliding to the right on the truck, were it not for F counteracting it. That is what P is meant to represent. I'm honestly not even sure if it's an actual force, but it is something for the friction to counteract so I thought it should be in the picture. Regarding the acceleration, I suppose there would be a centripetal acceleration going to the left in the picture, towards the centre of the curve. There is no mention of any tangential acceleration of the truck in the problem. Also yeah I definitely drew the N with a scuffed angle here...
 
  • #6
jack action
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Forget about ##P## for now. Show us how the centripetal acceleration would affect your FBD.
 
  • #7
jack action
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To better understand the problem, separate the box from the truck bed, and do a FBD for the box and another one for the truck bed surface.
 
  • #8
BvU
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it is moving along the truck in the curve
Correct. To move in a circular path, a net force is needed. How much (in symbols, not in numbers) ? And in which direction ?

N is pointing perpendicular to the ground/the truck
Right. So better draw it that way too :smile:
 
  • #9
Crunge
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To better understand the problem, separate the box from the truck bed, and do a FBD for the box and another one for the truck bed surface.
Truck2.jpg

Ok so it should be something like this I suppose. The truck is in a circular motion with the centre of the curve to the left. Although both the velocity of the truck and radius of curve is unknown. EDIT: Wait I didn't see this message ok hold on separate fbds ok got it
 
  • #10
jack action
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How does ##a_c## affect your FBD?

Although both the velocity of the truck and radius of curve is unknown.
It doesn't have to be numbers:
Homework Statement:: A box with the mass m is carried by a truck. The truck is driving in a slanted curve with the speed v. The radius of the curve is ρ and its slope is at the angle θ. There is friction between the box and the truck, and the box is not moving. Calculate using the gravitational acceleration g = 9,806.
 
  • #11
Crunge
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How does ##a_c## affect your FBD?


It doesn't have to be numbers:
Ok so since the truck has acceleration towards the curve's centre, what happens in regards to the box is that the "flooring" beneath it is going left, and this should be why the friction has to prevent the box going right on the truck bed, i think. Maybe this is what is actually happening rather than that force P that I invented. Compared to the truck, the box has a in the opposite direction? The problem still remains of getting a concrete number for N. You are probably right that the exact numbers for v and the radius are not needed, but I still have to figure out to what extent the acceleration affects N. I have to dogsit now so this problem will have to be put on hold, but I think I'm getting somewhere hopefully.
 
  • #12
kuruman
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Compared to the truck, the box has a in the opposite direction?
Absolutely not. The box is at rest with respect to the truck which means that it has the same velocity and acceleration as the truck at all times. The problem clearly states "##\dots~##and the box is not moving." That means that the box is not moving with respect to the truck.
 
  • #13
jack action
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How about applying Newton's second law of motion to your FBD ...
 
  • #14
Crunge
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How about applying Newton's second law of motion to your FBD ...
Ok right so F = m*a. There is acceleration "to the left" and thus, the sum of the forces in this situation should be in that direction as well. m is known, a is currently not but should be equal to v^2 / ρ (also unknown), and F for the box, I believe should be the combination of the friction force's x-component and Ns x-component. Unless the acceleration for the box is parallell to the truckbed in which case N would be perpendicular and not have an x-component, but I don't think that's right. So I think I need to figure out those x-components.
 
  • #15
kuruman
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Unless the acceleration for the box is parallell to the truckbed in which case N would be perpendicular and not have an x-component, but I don't think that's right.
It is not right. The box is moving in a circle the plane of which is parallel to the ground. If that's the case, in what direction should the acceleration be? As you say, the net force in the direction of the acceleration is the sum of the x-components of the normal force and the force of friction, x being the horizontal direction.
 
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  • #16
Lnewqban
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Ok right so F = m*a. There is acceleration "to the left" and thus, the sum of the forces in this situation should be in that direction as well. m is known, a is currently not but should be equal to v^2 / ρ (also unknown), and F for the box, I believe should be the combination of the friction force's x-component and Ns x-component. Unless the acceleration for the box is parallell to the truckbed in which case N would be perpendicular and not have an x-component, but I don't think that's right. So I think I need to figure out those x-components.
The inlined truck could be at rest and a winch could be pulling that box horizontally: it is the same thing.

The force that is turning the truck’s bed from under the box that tries to keep moving on a straight line can be assumed as perfectly horizontal and of constant value, like if the truck keeps going around a roundabout of radius ρ at constant speed v.

If the angle of the slanted curve could be increased, it would reach a value for which no friction would be needed to prevent any slide of the box.
The value of your friction force must fill up the gap between that ideal angle and the actual one.
 
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  • #17
kuruman
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To @Crunge :
Let me summarize what is known or has been discovered so far.
  1. The problem is asking you to find the maximum value of the normal force.
  2. The acceleration is horizontal and two forces contribute to the net force in that direction, friction and the normal force. As you said, Nx + fx = max.
  3. The acceleration is centripetal and ax = v2/R.
  4. It follows that as the speed increases, say the driver is pressing on the accelerator, the speed and hence the acceleration must increase. Therefore the N and f must also increase for the turn to be negotiated at fixed radius and incline angle.
  5. Static friction cannot increase forever but has a maximum value ##\text{f}_{\text{s}}^{\text{max}}=\mu_sN.## When that value is exceeded, sliding occurs and all bets are off.
Conclusion: The maximum value of the normal force occurs when sliding is just about to happen.
Can you find it? Hint: Examine what happens in the vertical direction.
 
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  • #18
Crunge
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To @Crunge :
Let me summarize what is known or has been discovered so far.
  1. The problem is asking you to find the maximum value of the normal force.
  2. The acceleration is horizontal and two forces contribute to the net force in that direction, friction and the normal force. As you said, Nx + fx = max.
  3. The acceleration is centripetal and ax = v2/R.
  4. It follows that as the speed increases, say the driver is pressing on the accelerator, the speed and hence the acceleration must increase. Therefore the N and f must also increase for the turn to be negotiated at fixed radius and incline angle.
  5. Static friction cannot increase forever but has a maximum value ##\text{f}_{\text{s}}^{\text{max}}=\mu_sN.## When that value is exceeded, sliding occurs and all bets are off.
Conclusion: The maximum value of the normal force occurs when sliding is just about to happen.
Can you find it? Hint: Examine what happens in the vertical direction.
Yes! Exactly, yeah I finally got it. This was satisfying to figure out because at first glance there seems to be so much crucial info missing. Thank you for the help.
 
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  • #19
kuruman
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You could also use the horizontal equation to find the speed at which sliding will occur. However, to get a number for that, you will need the radius of the circle, but it's good practice anyway.
 

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