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

[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?

Homework Equations

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

 

[PeterO]

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?

Homework Equations

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.

 

[lec0x]

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

 

[lec0x]

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

 

[PeterO]

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!

 

[PeterO]

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

F_app the applied force in the direction of the handle.

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