# Forces and vectors: pulling a baby buggy on soft ground

• Jeff97
In summary, the buggy is being pulled with a force of 100N at an angle of 40degrees to the horizontal. The vertical component of that force is 64.28N.
Jeff97
Homework Statement
A buggy with a child has a mass of 25kg. On the soft ground, it's easier to pull the buggy than push it. The buggy is pulled with a force of 100N at an angle of 40degrees to the horizontal.

Calculate the vertical component of the force and then us it to find the total force exerted by the wheels on the ground.
Relevant Equations
I don't know exactly where to start with this problem. But I'm going to try this, cos θ= adj/hyp cos 40° = Fx/100 Fx=100 cos 40° =76.60= 77N??

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Start by drawing a picture. Then draw a free body diagram showing all forces and angles and a coordinate system. Then do vector algebra.
Where does 42 come from?

How does this relate to the problem??

Yes, I'm sorry got two different problems confused, edited original post.

Jeff97 said:
Problem Statement: A buggy with a child has a mass of 25kg. On the soft ground, it's easier to pull the buggy than push it. The buggy is pulled with a force of 100N at an angle of 40degrees to the horizontal.

Calculate the vertical component of the force and then us it to find the total force exerted by the wheels on the ground.

Fx=100 cos 40° =76.60= 77N??
That's the horizontal component of the applied force, but you are not asked for that.

64.2516?

Units please and what is this exactly? (I think you are on the right track)

It's the vertical component. V = cos(θ) × f which is: V=sin40°×100N = 64.27=4N?

Jeff97 said:
It's the vertical component. V = cos(θ) × f which is: V=sin40°×100N = 64.27=4N?
64.28N. What is the "=4"?

typo 64N?

Jeff97 said:
typo 64N?
Ok. So on to the next part, what is the total force exerted on the ground?

@haruspex For the total force do I need to know acceleration? so do I need to do a=f/m?

Jeff97 said:
@haruspex For the total force do I need to know acceleration? so do I need to do a=f/m?
Excellent question. In principle, yes, but you have no way to determine it, so you will have to assume there is no acceleration.

This may sound a little stupid, but I'm not aware of that formula.

Jeff97 said:
This may sound a little stupid, but I'm not aware of that formula.
What formula?
If the buggy is not accelerating then you can consider the force balance on it.

I just don't know how to progress, I see your comment but don't quite understand it. Are you saying that the mass of the buggy is also the same force being pushed up at the ground?

Jeff97 said:
I just don't know how to progress, I see your comment but don't quite understand it. Are you saying that the mass of the buggy is also the same force being pushed up at the ground?
You mean the weight of the buggy, not the mass.
Have you drawn an FBD of the buggy? What forces act on it?
Since we are assuming it is not accelerating, what equations can you write?

F=mg? I have not drawn an FBD but I think there's the pull force to the right, then drag to the left, weight acting down and reactive force upwards

Jeff97 said:
F=mg? I have not drawn an FBD but I think there's the pull force to the right, then drag to the left, weight acting down and reactive force upwards
Yes, but you can think of the drag and normal force from the ground combined as a single total force from the ground.

hutchphd said:
Start by drawing a picture. Then draw a free body diagram showing all forces and angles and a coordinate system. Then do vector algebra.
Jeff97 said:
F=mg? I have not drawn an FBD but I think there's the pull force to the right, then drag to the left, weight acting down and reactive force upwards

Why should we bother?

@hutchphd, F=mg, mass is 64x9.8=627.2N this is the total force exerted?

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Where did the 64 come from? Certainly not the 64 N in frame #8.
In which direction are you now trying to work out the total force -- total force exerted on the ground?

Jeff97 said:
I have not drawn an FBD
Then do so, and you deserve no further assistance until then.
According to Newton, the total force the buggy exerts on the ground is equal and opposite to what?

Ok,@haruspex I have "drawn" one. So what your saying is the force being exerted by the buggy, the ground will exert the same amount of force in the opposite direction.(up)In my FBD, I have Identified A pulling force of 100N, downward force,weight, of 25kg and a normal force acting up.

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Jeff97 said:
Ok,@haruspex I have "drawn" one. So what your saying is the force being exerted by the buggy, the ground will exert the same amount of force in the opposite direction.(up)In my FBD, I have Identified A pulling force of 100N, downward force,weight, of 25kg and a normal force acting up.
With one force straight up, one straight down and one up at an angle it can't be constant velocity, can it? What have you forgotten? Hint: reread my post #19.

drag, Is there a formula for this?

Jeff97 said:
drag, Is there a formula for this?
You do not need a formula for it; since there is no acceleration you can deduce it from the other forces.
As I wrote in post #19, you do not need to think of drag and normal force separately. Just treat that combination as the total reaction force from the ground. That leaves you with only that unknown force and two known forces.

haruspex said:
Yes, but you can think of the drag and normal force from the ground combined as a single total force from the ground.
Ok, Is the unknown force gravity? or mg? (as I called the weight downwards) Deduction, interesting, I assume I need to list my force/s numerical values. Weight down (=mg= 25x9.8=245N) Pull force (100N) ?

Jeff97 said:
Is the unknown force gravity? or mg?
How is that unknown? You know the mass and you know g, so the weight is known.
Jeff97 said:
I need to list my force/s numerical values
And their directions. Forces are vectors, directions matter.
Jeff97 said:
Weight down (=mg= 25x9.8=245N) Pull force (100N) ?
Yes. The third force is the one of unknown value. What name did I give it in post #27?

"total reaction force downwards"

Jeff97 said:
"total reaction force downwards"
I do not see that phrase in post #27. What did I write? Do you understand the differences?

## 1. What is the difference between a force and a vector?

A force is a push or pull that acts on an object, while a vector is a quantity that has both magnitude and direction. In other words, a force is a type of vector.

## 2. How does pulling a baby buggy on soft ground affect the forces acting on the buggy?

Pulling a baby buggy on soft ground increases the amount of friction between the wheels and the ground, which creates a resistance force that must be overcome by the pulling force. This results in a net force acting on the buggy in the direction of motion.

## 3. How can we calculate the force required to pull a baby buggy on soft ground?

The force required to pull a baby buggy on soft ground can be calculated using the formula F = μmg, where F is the force, μ is the coefficient of friction, m is the mass of the buggy, and g is the acceleration due to gravity. The coefficient of friction can be determined by conducting experiments or consulting reference tables.

## 4. How does the angle of the pulling force affect the motion of the baby buggy?

The angle of the pulling force affects the motion of the baby buggy by changing the direction of the net force acting on the buggy. If the pulling force is applied at an angle, it will have both a horizontal and vertical component. The horizontal component will contribute to the forward motion of the buggy, while the vertical component will affect the weight of the buggy and the normal force exerted by the ground.

## 5. Can the weight of the baby affect the force needed to pull the buggy on soft ground?

Yes, the weight of the baby can affect the force needed to pull the buggy on soft ground. The greater the weight of the baby, the greater the force needed to overcome the resistance force of friction and move the buggy forward. This is why it may be more difficult to pull a heavier baby in a buggy on soft ground compared to a lighter baby.

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