# Find the missing force when the boy is pulled

• issacnewton
In summary: F_1## and ##F_2## are forces in the horizontal direction. So, we can use Newtons law to find the magnitude and direction of the father's force.$$F_1 \cos(\theta) + F_2 \cos(35) = ma$$$$F_1 = 100 N + \cos(35)$$$$F_2 = 100 N - \sin(35)$$$$F_1 \cos(\theta) + F_2 \cos(35) = 100 N$$$$F_1 = 100 N + 8.5 m/s^2\\$$F_2 = 100 N - 8.
issacnewton

## Homework Statement

A 15.0 kg boy sits on a 2.00 kg stake board and is pulled forward by his parents. His mom pulls with a force of 100. N at an angle of 35 degrees and the boy and skate board accelerate along the x-axis at 8.50m/s2? a. What is the magnitude and direction of the father’s force if the boy and skate board move along the x axis? b. How far would they get if they started from rest and the parents pulled for 4.00s?

Newton's law

## The Attempt at a Solution

Let ##F_2 = 100 N## be the force exerted by the Mom and ##F_1## be the force exerted by the dad and let ##\theta## be the angle made by this with the horizontal. Let ##m## be the mass of the boy + stake board. The we have weight ##mg## and the normal reaction upward ##N## acting on boy+board system. We can set up the following equatons.
$$F_1 \cos(\theta) + F_2 \cos(35) = ma$$
$$0 = N + F_2 \sin(35) + F_1 \sin(\theta) - mg$$
Now ##a## is also given, so we have 2 equations and three unknowns, ##F_1##, ##\theta##, and ##N##. So how would we solve this ?

This problem only makes sense to me if the x and y axes are two horizontal axes. We aren't considering vertical components at all. The father's vertical component could be anything that's not enough to accelerate the boy upward. Thus, the angles are horizontal angles relative to the forward direction, not vertical angles.

So under that assumption, ##F_2 \sin(35) + F_1 \sin(\theta) = 0##.

so what happens to ##N## and ##mg##. That is also part of the free body diagram.

N would be equal to mg.

Orodruin, why would N = mg ? Because of ##F_1## and ##F_2##, the normal reaction is not necessarily equal to the weight.

Because the problem disregards vertical forces completely, as stated already in #2. The forces applied by the parents are to be assumed to be horizontal.

but force applied by the mom is at 35 degrees with the horizontal and question is asking to find father's force magnitude and its direction. So if we assume that parents forces are horizontal, then question does not make sense.

IssacNewton said:
but force applied by the mom is at 35 degrees with the horizontal
No it is not. This is your interpretation (in which the problem statement does not make sense) not the statement of the problem. The formulation that makes sense is that the 35 degrees is 35 degrees relative to the direction of acceleration, but still in the horizontal plane.

acceleration is in the positive x direction. So you are right that ##F_1## makes angle 35 with the horizontal direction of the acceleration a. And that's what I was saying when I said that mom's force is at 35 degrees with the horizontal.

IssacNewton said:
acceleration is in the positive x direction. So you are right that ##F_1## makes angle 35 with the horizontal direction of the acceleration a. And that's what I was saying when I said that mom's force is at 35 degrees with the horizontal.
There are two horizontal directions. One which is the direction of the acceleration and one more, which is perpendicular to it.

ok, I take upward direction here as vertical direction and not the horizontal direction. So are you and me saying the same thing ?

No. In total there are three directions. Two horizontal and one vertical. In this problem there is no force component in the vertical direction.

Ok, let me put a free body diagram, otherwise we will be talking past each other. Is that FBD correct ?

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IssacNewton said:

## Homework Statement

A 15.0 kg boy sits on a 2.00 kg stake board and is pulled forward by his parents. His mom pulls with a force of 100. N at an angle of 35 degrees and the boy and skate board accelerate along the x-axis at 8.50m/s2? a. What is the magnitude and direction of the father’s force if the boy and skate board move along the x axis? b. How far would they get if they started from rest and the parents pulled for 4.00s?

Newton's law

## The Attempt at a Solution

Let ##F_2 = 100 N## be the force exerted by the Mom and ##F_1## be the force exerted by the dad and let ##\theta## be the angle made by this with the horizontal. Let ##m## be the mass of the boy + stake board. The we have weight ##mg## and the normal reaction upward ##N## acting on boy+board system. We can set up the following equatons.
$$F_1 \cos(\theta) + F_2 \cos(35) = ma$$
$$0 = N + F_2 \sin(35) + F_1 \sin(\theta) - mg$$
Now ##a## is also given, so we have 2 equations and three unknowns, ##F_1##, ##\theta##, and ##N##. So how would we solve this ?

I would stick with the first equation. Remember that acceleration on the right hand side is that caused by the net force, which in this case is in the x direction.

bluejay27 said:
I would stick with the first equation. Remember that acceleration on the right hand side is that caused by the net force, which in this case is in the x direction.
You should separate by components.

bluejay27 said:
You should separate by components.
By components, I mean in the x and y direction.

bluejay27, I have two equations in my original post. One for x direction and other for y direction. So ##F_1 \cos(\theta)## would be the x component of the force ##F_1##. I realize that I have interchanged the forces ##F_1## and ##F_2## in the free body diagram. Sorry about that.

bluejay27
IssacNewton said:
Ok, let me put a free body diagram, otherwise we will be talking past each other. Is that FBD correct ?
No. You have drawn vertical force components. There are no vertical force components apart from the gravitation and the normal force.

Orodruin, are you saying that both the forces are acting along the positive x direction ? Then what is 35 degrees referring to ?

IssacNewton said:
bluejay27, I have two equations in my original post. One for x direction and other for y direction. So ##F_1 \cos(\theta)## would be the x component of the force ##F_1##. I realize that I have interchanged the forces ##F_1## and ##F_2## in the free body diagram. Sorry about that.
I would just solve for the force of the dad. Since the sum of the forces are already restricted in the direction of x. For the magnitude and direction of the force of the dad, you can now use the sum of the forces in the y direction.

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of course the motion is only in the horizontal direction, but ##F_2## has component in the vertical direction. And we can't assume that ##F_1## is acting only in the forward direction.

IssacNewton said:
Orodruin, are you saying that both the forces are acting along the positive x direction ? Then what is 35 degrees referring to ?
No. Again, the x direction is not the only horizontal direction.

IssacNewton said:
of course the motion is only in the horizontal direction, but ##F_2## has component in the vertical direction. And we can't assume that ##F_1## is acting only in the forward direction.
Solve for the unknown force of x. Then solve for the unknown force of y. Only use the mass of the boy when solving in the y direction. You will then be able find the magnitude and angle with Fx and Fy of the dad by forming a triangle.

IssacNewton said:
so what happens to ##N## and ##mg##. That is also part of the free body diagram.

If we assume that ##F_1## and ##F_2## are in the horizontal plane, then ##N = mg## and is irrelevant.

IssacNewton said:
Orodruin, why would N = mg ? Because of ##F_1## and ##F_2##, the normal reaction is not necessarily equal to the weight.
Yes, it would because we are saying that there is no vertical component of ##F_1## and ##F_2##. If ##x## is East, then the mother is pulling northeast and the father is pulling southeast.

The problem didn't say that explicitly, didn't say whether those angles are vertical angles or horizontal ones. But for the reasons you gave, the problem doesn't make sense if they are vertical angles. As I said in my initial response, the problem only makes sense if all the forces and angles are in the horizontal plane.

Orodruin is completely right. If we are to solve this confusion, we must realize that:
1. There are three directions, which I will call here, x, y (both in the horizontal plane) and the vertical direction, which we can call z.
2. The original statement of the problem does not make point 1 clear, which is the driving force behind the confusion.
3. The problem can be solved if we follow what Orodruin is saying. Both forces are in the horizontal plane. The angles given are therefore angles in the horizontal plane. N = mg will then be correct, but irrelevant to the problem because both N and mg are along the vertical (z) direction.

Imagine the boy is in a harness with a loop onto which the pulling ropes are attached so that Mum can pull one rope and Dad the other. Mum's rope is at 35 degrees to the direction of motion and Dad's at unknown θ. The cos components of the rope tensions provide the forward force whilst the sine components presumably cancel so that the boy does not move sideways. Both ropes are in the same horizontal plane as the loop to which they are attached. I think this is what Oordruin is indicating when he says there are no vertical force components other than mg and N.

RPinPA said:
If ##x## is East, then the mother is pulling northeast and the father is pulling southeast.

Oh. Now I see what you mean. Now if I solve it, then I get ##\theta = 35## degrees and dad's force is ##F_1 = F_2 = 100 N##. Is that right ?.
But the problem is really confusing. It should explicitly tell the student what the question is talking about.

What if the mother - also a superhero - doesn't want her child to get windburn, and is actually pulling backwards against the father's overenthusiasm, that ends up pulling them along at ##8m/s^2## for 4 seconds.

Make reasonable assumption(s), state them, then you can solve the problem.

In this case the reasonable assumptions for a mildly boring answer are that the parents' pulls are both positive along the ##x## axis, cancel each other out on the ##z## axis, and don't exceed the force of gravity acting on the child, on the ##y## axis. Also, the ##35°## of the mother's pull is off the x-axis.

IssacNewton said:
Now if I solve it, then I get θ=35\theta = 35 degrees and dad's force is F1=F2=100NF_1 = F_2 = 100 N. Is that right ?

And that doesn't strike you as being a little bit too easy ?

Okay, ignore the possible ##y## axis component (though you really should do that for fun, separately), and assume that both pulls are entirely on the ##xz## (horizontal) plane.

Show your work.

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IssacNewton said:
Orodruin, are you saying that both the forces are acting along the positive x direction ? Then what is 35 degrees referring to ?

Let me try. Picture a coordinate system with all three axes x, y, and z. X points east, y points north, and z points up. What Orodruin is trying to tell you is that the problem is in the x y plane. There is no up. The angle refers to the angle from the x axis. Mom is pulling North of east. In order for the boy to move East Dad must pull south of east. The two dimensions give you two equations and so you can solve for both the magnitude and direction of Dad’s force.

Regarding N=mg, that does not act in the x y plane. Also skateboard = “no friction”, so all you have are mom’s and dad’s forces and inertia.

The skateboard as seen from above. It is rolling in the plane of the screen. The two forces are also in the plane of the screen and have no component perpendicular to the screen. The x-axis is horizontal. A picture is worth 210 words.

On Edit: In part (b) we are told that they accelerate for 4 s. Starting from rest at an acceleration of 8.5 m/s2, they would reach a speed of 34 m/s or 76 miles per hour! They must be using sports cars to pull the kid which constitutes reckless endangerment and child abuse. Don't ry this at home.

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Ok, I think I did wrong calculation in post 28. We have $$F_2 \cos(35) + F_1\cos(\theta) = ma$$ and we also have ##F_2\sin(35) = F_1\sin(\theta)##. This together leads to $$\tan(\theta) = \frac{F_2\sin(35)}{ma - F_2 \cos(35)}$$ Plugging ##m = 17## kg and ##a= 8.5\; m/s^2##, we would get ##\theta = 42.5## degrees. And then ##F_1 = F_2\sin(35)/\sin(\theta) = 84.9 N##. I hope this is correct.

Your answer for part (a) agrees with mine.

IssacNewton said:
But the problem is really confusing. It should explicitly tell the student what the question is talking about.
My view is that the problem is well formulated (it is a VERY common type of introductory mechanics problem) but you have given it a very weird and unexpected interpretation that is totally different than what just about everyone else would ever give it. When you have encountered a few more of these problems you'll see what I mean.

phinds said:
My view is that the problem is well formulated (it is a VERY common type of introductory mechanics problem) but you have given it a very weird and unexpected interpretation that is totally different than what just about everyone else would ever give it. When you have encountered a few more of these problems you'll see what I mean.
My view differs somewhat from yours. I agree that, after seeing several of these problems, one acquires an intuitive understanding of how to interpret them. That's a sign that one has made the transition from novice to expert. However, I think that the formulator of a well crafted problem should look ahead for possible misunderstandings of the wording and include a figure if there is a possibility of misinterpretation by novices. I also think that the given acceleration is totally unrealistic and absurd. Numerical answers should provide an understanding and an estimating ability of the numbers associated with the quantities that are used. The formulator of this problem apparently did not pay attention to that aspect. For these reasons, I would say that this problem is so-so formulated.

<h2>1. What is the missing force when the boy is pulled?</h2><p>The missing force when the boy is pulled is the force required to move the boy in the direction of the pull. It is the force that must be applied to overcome the resistance or inertia of the boy.</p><h2>2. How is the missing force calculated?</h2><p>The missing force can be calculated using the formula F = ma, where F is force, m is mass, and a is acceleration. The mass of the boy and the acceleration at which he is being pulled must be known in order to calculate the missing force.</p><h2>3. What factors affect the missing force when the boy is pulled?</h2><p>The missing force is affected by the mass of the boy, the acceleration at which he is being pulled, and any other forces acting on the boy such as friction or air resistance. The angle and direction of the pull also play a role in determining the missing force.</p><h2>4. How does the missing force affect the boy's movement?</h2><p>The missing force is directly proportional to the acceleration of the boy. This means that the greater the missing force, the faster the boy will move in the direction of the pull. If the missing force is too small, the boy may not move at all or may move very slowly.</p><h2>5. Can the missing force be negative?</h2><p>Yes, the missing force can be negative if the boy is being pulled in the opposite direction of his current movement. This can happen if there is a force acting on the boy in the opposite direction, such as friction or air resistance. In this case, the missing force would be a resistance force, rather than a force that causes movement.</p>

## 1. What is the missing force when the boy is pulled?

The missing force when the boy is pulled is the force required to move the boy in the direction of the pull. It is the force that must be applied to overcome the resistance or inertia of the boy.

## 2. How is the missing force calculated?

The missing force can be calculated using the formula F = ma, where F is force, m is mass, and a is acceleration. The mass of the boy and the acceleration at which he is being pulled must be known in order to calculate the missing force.

## 3. What factors affect the missing force when the boy is pulled?

The missing force is affected by the mass of the boy, the acceleration at which he is being pulled, and any other forces acting on the boy such as friction or air resistance. The angle and direction of the pull also play a role in determining the missing force.

## 4. How does the missing force affect the boy's movement?

The missing force is directly proportional to the acceleration of the boy. This means that the greater the missing force, the faster the boy will move in the direction of the pull. If the missing force is too small, the boy may not move at all or may move very slowly.

## 5. Can the missing force be negative?

Yes, the missing force can be negative if the boy is being pulled in the opposite direction of his current movement. This can happen if there is a force acting on the boy in the opposite direction, such as friction or air resistance. In this case, the missing force would be a resistance force, rather than a force that causes movement.

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