# Find the missing force when the boy is pulled

## 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 ?

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RPinPA
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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.

Orodruin
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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.

Orodruin
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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.

Orodruin
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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.

Orodruin
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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 ?

Orodruin
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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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## 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.

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.

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
Orodruin
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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 ?

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.

Orodruin
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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.

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.

RPinPA
Homework Helper
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.

RPinPA
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.