# What Is the Acceleration of a Wagon Pulled at an Angle?

• lec0x
In summary, a child pulling a wagon with a force of 59 N at a 20 degree angle with the horizontal causes an acceleration of 10.66 m/s2 for the wagon, which has a mass of 5.2 kg. The free body diagram for this situation includes a weight force (mg), a normal reaction force (N), and an applied force (Fapp).
lec0x
Force With Angles!

## Homework Statement

A child pulls a wagon with a force of 59 N by a handle making an angle of 20 degrees with the horizontal. If the wagon has a mass of 5.2 kg, to the nearest hundredth of a m/s2 what is the acceleration of the wagon?

ƩF=ma

## The Attempt at a Solution

I drew a picture of the wagon and I know that if a child is exerting a force of 59N on the handle then the handle is exerting that back. does that mean that the handle is exerting that force onto the wagon as well?
The free body diagram has Force Normal going up, mgsin(θ) to the right as well as Force Applied, mgcos(θ) going down and Force friction going to the left!

ƩFy=ForceN-mgcos(θ)=0 so FN=mgcos(20)
and then ƩFx = Fpp+mgsin(20)-umgcos(20)
then would the force applied just be 59N or would there have to be an equation using θ

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lec0x said:

## Homework Statement

A child pulls a wagon with a force of 59 N by a handle making an angle of 20 degrees with the horizontal. If the wagon has a mass of 5.2 kg, to the nearest hundredth of a m/s2 what is the acceleration of the wagon?

ƩF=ma

## The Attempt at a Solution

I drew a picture of the wagon and I know that if a child is exerting a force of 59N on the handle then the handle is exerting that back. does that mean that the handle is exerting that force onto the wagon as well?
The free body diagram has Force Normal going up, mgsin(theta) to the right as well as Force Applied, mgcos(theta) going down and Force friction going to the left!

ƩFy=ForceN-mgcos(theta)=0 so FN=mgcos(20)
and then ƩFx = Fpp+mgsin(20)-umgcos(20)
is this right so far?

I reckon you have you sin and cosine backwards, and I don't see the whole weight involved. Given that you have not given a co-efficient of friction is that actually involved. The wagon probably has wheels, and the purpose of the wheels is to [all buy] remove the friction from the situation.

I just remember in class my teacher saying that x is sin and y is cosine even though it's usually the other way. so there is no μ and no friction, would the force applied be on the opposite side of Fgx so it would be
Fapp-mgcosθ=ma
59-(5.2)(9.8)cos(20)=(5.8)a
11.11=5.8a
a=1.92

PeterO said:
I reckon you have you sin and cosine backwards, and I don't see the whole weight involved. Given that you have not given a co-efficient of friction is that actually involved. The wagon probably has wheels, and the purpose of the wheels is to [all buy] remove the friction from the situation.

nevermind i figured it out you would do
59cos(20)=55.4
then you would do
55.4=ma
55.4=(5.2)a
a=10.66

lec0x said:
nevermind i figured it out you would do
59cos(20)=55.4
then you would do
55.4=ma
55.4=(5.2)a
a=10.66

Now you have it!

lec0x said:
nevermind i figured it out you would do
59cos(20)=55.4
then you would do
55.4=ma
55.4=(5.2)a
a=10.66

BTW, Your free body diagram should have shown 3 forces only.

mg the weight down.

N the normal reaction Force Up

Fapp the applied force in the direction of the handle.

That Fapp would be resolved into appropriate components to get the acceleration, as you did.

## 1. What is the magnitude of the force applied by the child?

The magnitude of the force applied by the child is 59 N.

## 2. What is the direction of the force applied by the child?

The direction of the force applied by the child is 20 degrees from the horizontal.

## 3. How is the force related to the angle of the handle?

The force applied by the child is directly proportional to the angle of the handle. As the angle increases, the force also increases.

## 4. Will changing the angle of the handle affect the force applied by the child?

Yes, changing the angle of the handle will affect the force applied by the child. The force will increase as the angle increases and decrease as the angle decreases.

## 5. How do I calculate the horizontal and vertical components of the force?

You can use trigonometric functions (sine, cosine, and tangent) to calculate the horizontal and vertical components of the force. The horizontal component can be calculated by multiplying the magnitude of the force by the cosine of the angle, while the vertical component can be calculated by multiplying the magnitude of the force by the sine of the angle.

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