Acceleration of slinding crate pushed and pulled at the same time

[acherentia]

Two workers must slide a crate designed to be pushed and pulled at the same time as shown in the figure below. Joe can always exert twice as much force as Paul, and Paul can exert 155 N of force. The crate has a mass m = 41 kg, the angle θ = 20° and the coefficient of kinetic friction between the floor and crate is 0.55.

http://s783.photobucket.com/albums/yy113/eandronic/?action=view&current=4-p-073.gif

If we want to move the crate as fast as possible, is it better to have Paul push and Joe pull, or vice-versa?
Intuitively, I consider it's better to have Paul push, and Joe pull.

Calculate the (horizontal) acceleration for both cases to find out.

Paul pushing m/s2 ?
Joe pushing m/s2 ?

I am not sure which are all the forces on the x-axis acting on the crate and how to calculate them at the given angle?

 

[Delphi51]

What exactly is the question?
It will take a long time to get your attachment approved so we can see it. Suggest you upload the diagram to a free photo site like photobucket.com and post a link here.

 

[acherentia]

What exactly is the question?
It will take a long time to get your attachment approved so we can see it. Suggest you upload the diagram to a free photo site like photobucket.com and post a link here.

http://s783.photobucket.com/albums/yy113/eandronic/

If we want to move the crate as fast as possible, is it better to have Paul push and Joe pull, or vice-versa?
1 It's better to have Paul push, and Joe pull.

Calculate the (horizontal) acceleration for both cases to find out.

Paul pushing m/s2 ?
Joe pushing m/s2 ?

 

[Delphi51]

Okay, so you'll have to use trig to find the horizontal and vertical parts of each force in each of the two scenarios. And find the friction force. Then you'll be in a position to do "sum of forces = ma".

 

[acherentia]

what i tried is
when paul pushes:
a(x)=fpush_paul*cos(180-20)+fpull_joe*cos20 -Ff

when joe pushes:
a(x)=fpush_joe*cos160+f pull_paul*cos20 - Ff

and I am not sure of this and I also don't know the correct expression for Fn, therefore I don't have the good expression for Ff.

in both cases i get negative accelerations.

 

[Delphi51]

You have forgotten to include friction.
Careful with cos(180-20). It is negative. Better to use cos(20) I'd say.

 

[acherentia]

I have included and subtracted friction. Just that I don't know the formula for it.

Ff=Fn*miu(k)

miu of kinetic friction is given, Fn is unknown and I am not sure how to express it.

One option is Fn - m*g*sin_20 + Fpull_sin_20+ Fpush_sin_20 = 0 the other option is Fn=mgsin20.

Regarding the acceleration on the x axis:
If the angle is 20 in both cases would this be a right formula then? I am not sure what it is that quantifies the difference that I intuitively understand should give me a greater magnitude when the stronger guy pulls.

when paul pushes:
a(x)=(fpush_paul*cos20+fpull_joe*cos20 -Ffriction) / mass

when joe pushes:
a(x)=(fpush_joe*cos20+f pull_paul*cos20 - Ffriction) / mass

Please advise.

 

[Delphi51]

Yes, less friction when Joe pulls because the stronger force is partly upward and it therefore reduces Fn.
Fn is the total vertical (downward) force pushing the block against the ground. Add mg, the component of Joe's force that is upward and the component of Paul's force that is downward.

 

[madz89]

ok..note :paul=155N and joe=310N
case 1:when paul pushes and joe pulls..
ma(y)=N+F_joe sin20-F_paul sin20 - mg
a(y)=0; solve for N
ma(x)=F_joe cos20+F_paul cos20-miu N

case 2:when joe pushes and paul pulls
ma(y)=N+F_paul sin20-F_joe sin20 - mg
a(y)=0 ;solve for N
ma(x)=F_joe cos20+F_paul cos20-miu N