Finding the Magnitude of Pushing Force with Kinetic Friction?

[jr4life]

Homework Statement

The block is pushed to the right at a constant velocity. If the coefficient of kinetic friction is 0.5, what is the magnitude of the pushing force?

Homework Equations

I don't know...

The Attempt at a Solution

 

[learningphysics]

Did they give the mass of the block? What's your first idea? Give it a shot...

 

[jr4life]

no they didnt give the mass of the block, that is why I am so confused my teacher said something about sigma Fx=0 and sigma Fy=0 but i don't know what to do with those numbers, all i know is velocity is constant which makes acceleration 0 which makes force of x and y 0, but after that I am lost

 

[learningphysics]

no they didnt give the mass of the block, that is why I am so confused my teacher said something about sigma Fx=0 and sigma Fy=0 but i don't know what to do with those numbers, all i know is velocity is constant which makes acceleration 0 which makes force of x and y 0, but after that I am lost

Yes, net force in the x direction is 0. Net force in the y-direction is 0.

What are the forces acting in the y-direction?

What are the forces acting in the x-direction?

 

[jr4life]

In the x direction is the pushing force and force of friction and in the y direction is force of gravity on a box whos mass is not given and the normal force, i think the lack of given numbers is confuing me

 

[learningphysics]

In the x direction is the pushing force and force of friction and in the y direction is force of gravity on a box whos mass is not given and the normal force, i think the lack of given numbers is confuing me

Yeah, just use the variables... let mass = m.

Write the \Sigma\vec{F} = ma equation for the x-direction... then for the y-direction... just use the variables...

 

[jr4life]

in x direction it would be (mass)(0) so resultant force would be 0, because acceleration would be 0 due to constant velocity and in y direction (mass)(9.8) and then i don't really know what you could do with that...

 

[learningphysics]

in x direction it would be (mass)(0) so resultant force would be 0, because acceleration would be 0 due to constant velocity and in y direction (mass)(9.8) and then i don't really know what you could do with that...

Write the equations using Fpushing, friction, mg, Fnormal... how do they add or subtract?

 

[jr4life]

Write the equations using Fpushing, friction, mg, Fnormal... how do they add or subtract?

mg-Fnormal=0

this is where i get confused with x because they are not in equilibrium so they wouldn't equal 0
Fpushing-friction=?? idk

 

[learningphysics]

mg-Fnormal=0

this is where i get confused with x because they are not in equilibrium so they wouldn't equal 0
Fpushing-friction=?? idk

It equals 0, because acceleration is 0 (constant velocity). Now, you also know that friction = \mu * F_{normal}

Solve for Fpushing using your two equations, in terms of mass, \mu and g.

 

[jr4life]

It equals 0, because acceleration is 0 (constant velocity). Now, you also know that friction = \mu * F_{normal}

Solve for Fpushing using your two equations, in terms of mass, \mu and g.

ok, i understand why it equals zero now but when you use that equation would it be
friction=(0.5)(9.8m) and which two equations are you referring to?

 

[learningphysics]

ok, i understand why it equals zero now but when you use that equation would it be
friction=(0.5)(9.8m) and which two equations are you referring to?

Yes, that's right...

The two equations I meant were:

Fpushing - friction = 0
Fnormal - mg = 0

you actually used the second equation to get friction = 0.5*9.8*m

So what does Fpushing come out to...

 

[jr4life]

Yes, that's right...

The two equations I meant were:

Fpushing - friction = 0
Fnormal - mg = 0

you actually used the second equation to get friction = 0.5*9.8*m

So what does Fpushing come out to...

Fpushing-(0.5)(9.8m)=0
Fpushing-4.9m=0
Fpushing=4.9m??

 

[FedEx]

Fpushing-(0.5)(9.8m)=0
Fpushing-4.9m=0
Fpushing=4.9m??

Seems good to me.

 

[learningphysics]

Seems good to me.

yup. looks good to me too.