# Using Advanced Dynamics to find the Force Applied of a Truck

• QuickSkope
In summary: CosθSolving for a, we get a = -gTanθBut now how do we get Fa? I'm not sure what you mean. You have a = -gTanθ as the acceleration of the block, so Fnet = ma gives you the horizontal component of the normal force. Isn't that what you wanted? Then you just need to relate that to the force of the truck by considering the forces on the truck, ignoring its weight (it's not moving vertically) to find the force it exerts.In summary, the normal force on the block can be broken down into horizontal and vertical components, and by considering the forces on the block and the truck, we can find
QuickSkope

## Homework Statement

A car is moving at an acceleration such that a 5kg block in the back of the remains still on a ramp at 35*. The ramp is frictionless, whereas the coefficient of friction between the truck and the road is μ=0.1. If the truck weighs 12000kg, what is the force the trucks engine is exerting?

Dynamics formula
Fnet Formulas

## The Attempt at a Solution

We know that the Ramp only has Force Parrallel pushing it down.

Parrallel= mgSinθ
= (9.8)(5)(Sin35)
=28.105

We also know that the only thing that would push the block UP the ramp is the Fnet of the truck. For the Block to remain still, it has to have an Fnet of 0. Since its on a Ramp, you have to find the Fx.

Fx= Cos35*28.105 = 23.02.

From there, we know the Fnet of the truck is 23.02.

Fnet = Fa - Ff
23.02= Fa - (0.1)(9.8)(12000)
Fa = 11783.02

Does that seem right? I feel like it is a little bit off. Any help would be much appriciated.

Let the normal force on the block be N. Write out the ƩF=ma equations for the vertical and horizontal directions.

Dont you already know the normal force on the block? Its the Perpendicular Force right? (9.8*5*Cos35)

As for the Vertical and horizontal directions:

Fx=(9.8)(5)Sin35 = 28.105
Fy=(9.8)(5)Cos35 = 40.138

QuickSkope said:
Dont you already know the normal force on the block? Its the Perpendicular Force right? (9.8*5*Cos35)
That's the wrong value. Please write out the equations as I suggested.

My Calculations in the above post (#3) are off.

Fx= (28.105)Sin35 = 23.02
Fy= (28.105)Cos35 = 16.12

That might be better, cause those are the forces acting in the Y and X component (Fparallel broken up into X and Y components)

Am I on the right track?

EDIT: Okay, doing as you asked with the Fn.

Fx= FnSin35
Fy= FnCos35

Like that?

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QuickSkope said:
Fx= FnSin35
Fy= FnCos35
Yes, those are the horizontal and vertical components of the normal force. Now get each into a ƩF = ma equation by bringing in g and the known horizontal acceleration.

Fy=(9.8)(5)Cos35 = 40.1385

Is the known horizontal acceleration just (9.8)(Sin35)=5.6 m/s^2?

Thus the Fx= (5.6)(5) = 28.105?

I think I am wrong, as that's just the answer I got before.

EDIT: We don't really know the horizontal acceleration, as it is just stated that it is enough that the block appears motionless. From this, we know that the Fnet of the truck is the Fa on the block, Right?

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QuickSkope said:
Fy=(9.8)(5)Cos35 = 40.1385
In your previous post you wrote, correctly, that the vertical component of the normal force, Fn, is Fn cos(θ). (It's best to work entirely in symbols, only plugging in the numbers at the end.) The only other force on the block is gravity. What's the vertical component of that? What is the resulting vertical acceleration? Write an equation relating those three things.

The horizontal component of the normal force is Fn sin(θ). What other horizontal forces act on the block? Let the resulting horizontal acceleration on the block be a. Write the horizontal ƩF=ma equation.
We don't really know the horizontal acceleration, as it is just stated that it is enough that the block appears motionless. From this, we know that the Fnet of the truck is the Fa on the block, Right?
Yes, sorry, I was wrong to say the horizontal acceleration is known... that's part of what we're trying to determine.

It was my understanding that the only force that was acting down the ramp in this situation would be Force Parrallel (mgSinθ), is that not correct? (There is no friction on the ramp,only between the truck and the road.) After finding the force parralele, you could break it up into X and Y components and use that to find the Fa of the truck. I thought Fn would be irrelevant as there was no force of friction.Onto what you said, The verticle component of gravity is just mg, correct? The resulting verticle acceleration is just the addition of the Vectors of normal force and gravity. So that would mean the verticle acceleration would be a = ((FnCosθ) + mg) / 5)

For the horizontal component, there is also the Fnet of the truck pushing it towards the ramp. So the X component of the Fn is = to the Fnet of the truck.

Sorry if this is all wrong, I am very confused ATM. If you could answer the top in detail and explain what forces are pushing where, I would much appreciate it.

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QuickSkope said:
It was my understanding that the only force that was acting down the ramp in this situation would be Force Parrallel (mgSinθ), is that not correct? (There is no friction on the ramp,only between the truck and the road.) After finding the force parralele, you could break it up into X and Y components and use that to find the Fa of the truck.
Yes, that's the only force acting down the ramp, but we need the horizontal force, and the normal force is involved in that.
Onto what you said, The verticle component of gravity is just mg, correct? The resulting verticle acceleration is just the addition of the Vectors of normal force and gravity. So that would mean the verticle acceleration would be a = ((FnCosθ) + mg) / 5)
Right, and since there is no net vertical acceleration we can write FnCosθ = -mg
For the horizontal component, there is also the Fnet of the truck pushing it towards the ramp.
No, that's committing the sin of using a non-inertial frame. Don't worry about the truck at the moment. There is only one horizontal force on the block, Fn sin(θ), and that produces an acceleration a, so we have Fn sin(θ) = ma. Combine that with FnCosθ = -mg to eliminate Fn. What do you get?

Ahh, that makes more sense. So when we put the FnSinθ = ma and solve for Fn, we get Fn=ma/Sinθ

Doing the same with FnCosθ = -mg, we get Fn = -mg/Cosθ.

Equating the two, we get

ma/Sinθ = -mg/Cosθ

Now we have all the Values to solve for A.

(5)(a)/Sin35 = -(5)(-9.8)/Cos35

a = (49/Cos35) / (5/Sin35)

a = 6.86 m/s^2

Correct? That gives us our Acceleration in the X.

Expanding on that, we now know the acceleration in the X required to keep the block upwards, so that must be the acceleration of the truck. From there, we can calculate the Fnet of the Truck (6.86 * 12000) = 82344.41

Fnet = Fa - Ff
82344.41 = Fa - (9.8)(12000)(0.1)
Fa = 94104.41 N

I rounded when writing the numbers down here, but did the math without rounding in my calculator, so if youre doing the calculations I've written down, they might come out a tad off.

QuickSkope said:
ma/Sinθ = -mg/Cosθ
Right. And you can immediately simplify that to a = -g tan(θ). (The mass of the block doesn't matter.)
the Fnet of the Truck (6.86 * 12000) = 82344.41
Yes.
Fnet = Fa - Ff
No. It's a common error to think that the friction between the tyres and the road acts to slow a vehicle. That's only true when the vehicle is braking. When a vehicle is accelerating (positively), the friction provides that acceleration. (If there were no friction, the vehicle could not accelerate.) For a vehicle (or whatever) on a horizontal surface, if the only horizontal force acting is friction then frictional force = mass * acceleration.
The question asks for the force from the engine. That's rather an odd question. Strictly speaking, a car engine does not exert a linear force; it produces power and torque. The wheels convert the torque into force. How well that force converts to forward acceleration (as opposed to merely accelerating the spin of the wheels) depends on available friction.
I would give 82344N as the answer (or, more accurately, 82430N or 82400N, depending on the precision appropriate.)

haruspex said:
No. It's a common error to think that the friction between the tyres and the road acts to slow a vehicle. That's only true when the vehicle is braking. When a vehicle is accelerating (positively), the friction provides that acceleration. (If there were no friction, the vehicle could not accelerate.) For a vehicle (or whatever) on a horizontal surface, if the only horizontal force acting is friction then frictional force = mass * acceleration.
The question asks for the force from the engine. That's rather an odd question. Strictly speaking, a car engine does not exert a linear force; it produces power and torque. The wheels convert the torque into force. How well that force converts to forward acceleration (as opposed to merely accelerating the spin of the wheels) depends on available friction.
I would give 82344N as the answer (or, more accurately, 82430N or 82400N, depending on the precision appropriate.)

This seems odd to me, as that means the μ would be irrelevant. Your engine would have to generate a lot more force to get out of quicksand than accelerating on standard pavement, correct?

QuickSkope said:
This seems odd to me, as that means the μ would be irrelevant.
The relevance of the friction is to whether it is actually possible to keep the block stable on the ramp. Seems it is not. The maximum acceleration of the truck is gμ =0.98m/s2. But as you calculated, the acceleration needs to be 6.87m/s2.
Reading the question again, I see there are two things that are not clear. Which way is the ramp facing? If it faces backwards, the truck is decelerating, so it doesn't require the engine to be doing anything - it could be just on the brakes. But that doesn't get past the 0.1 friction problem. Secondly, it doesn't say the road is level. But if we don't know the angle of the road then we cannot answer the question.
Your engine would have to generate a lot more force to get out of quicksand than accelerating on standard pavement, correct?
Yes, but largely because getting out of quicksand involves going up as well as along. Also, spinning the wheels at a great rate to get out of sand means the propulsion is more like a jet than a normal rolling tyre; the forward force comes from the momentum of the sand thrown backwards.

The Ramp is indeed facing the same way the car is accelerating, and the road is level.

Making a little bit more sense, Thanks for everything!

Is there somewhere I can like give you a thumbs up or something?

QuickSkope said:
The Ramp is indeed facing the same way the car is accelerating, and the road is level.

Making a little bit more sense, Thanks for everything!
That's good, but it bothers me that the question no longer makes sense to me. The short answer is that with friction so low it is not possible for the block to stay in equilibrium on the ramp, no matter what the engine does. I'm going to post this thread to the homework helpers forum.
Is there somewhere I can like give you a thumbs up or something?
I believe there is, but I don't know how it's done. Thanks for the thought.

Hi QuickSkope!
QuickSkope said:
A car is moving at an acceleration such that a 5kg block in the back of the remains still on a ramp at 35*. The ramp is frictionless, whereas the coefficient of friction between the truck and the road is μ=0.1. If the truck weighs 12000kg, what is the force the trucks engine is exerting?

This has all been very complicated for a simple question.

As with most dynamics problems, you simply apply good ol' Newton's second law (F = ma) in the appropriate direction.

In this case (as you've correctly noticed) the appropriate direction is along the slope.

The only force (in that direction) is gravity.

Call the acceleration of the truck (or is it a car? ) "a".

Then what is the acceleration of the block? And what is the component of acceleration of the block along the slope?

Now do F = ma for the block to find a.

(Alternatively, have you done accelerating frames? If so, the effective gravity is the vector sum of the actual gravity and the fictitious force, and it has to be normal to the slope)

QuickSkope, I passed this problem around the team and the consensus is that although the question says the friction between the car and the road is 0.1, that's not what it means. Instead, they believe it is referring to rolling resistance. If so, your earlier treatment of it (adding to the force the engine needs to produce) is correct. Is it possible you copied this wording inaccurately? If not, I think it's a very badly worded and misleading question.
tiny-tim's solution is a little more direct, but - how can I put this? - I feel that you would do better sticking to a standard approach of resolving vertically and horizontally every time. Yes, it can be a bit more work, but you're less likely to outsmart yourself. OTOH, if you feel comfortable with tiny-tim's method, go right ahead.

## 1. What is Advanced Dynamics and how can it be used to find the force applied of a truck?

Advanced Dynamics is a branch of physics that studies the movement and behavior of objects under the influence of forces. By using equations such as Newton's Second Law and the Law of Conservation of Energy, we can calculate the force applied on a truck by analyzing its motion and acceleration.

## 2. What are some factors that can affect the force applied on a truck?

The force applied on a truck can be affected by a variety of factors such as the weight of the truck, its speed, the incline of the road, and any external forces acting on the truck (e.g. wind resistance). These factors can change the overall acceleration and movement of the truck, which in turn affects the force applied.

## 3. How do you measure the force applied on a truck using Advanced Dynamics?

To measure the force applied on a truck, we need to first collect data on its acceleration and mass. This can be done by using sensors and instruments attached to the truck or by recording its movement with cameras and analyzing the footage. Once we have this data, we can use equations from Advanced Dynamics to calculate the force applied.

## 4. Can Advanced Dynamics be used to find the force applied on a truck in real-time?

Yes, Advanced Dynamics can be used in real-time to continuously monitor and calculate the force applied on a truck. This is especially useful for industries such as transportation and logistics, where knowing the force applied on a truck can help improve safety and efficiency.

## 5. Are there any limitations to using Advanced Dynamics to find the force applied on a truck?

While Advanced Dynamics is a powerful tool for analyzing the force applied on a truck, it does have its limitations. Factors such as friction, air resistance, and other external forces can affect the accuracy of the calculations. Additionally, Advanced Dynamics calculations are based on ideal conditions and may not accurately reflect real-world scenarios.

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