Work done on crate moving on incline

[I_Try_Math]

Homework Statement

Shown below is a 40-kg crate that is pushed at constant velocity a distance 8.0 m along a incline by the horizontal force F. The coefficient of kinetic friction between the crate and the incline is u_k = 0.4 Calculate the work done by (a) the applied force, (b) the frictional force, (c) the gravitational force, and (d) the net force.

Relevant Equations

F = ma

For part a and b, I can't see a clear path to finding the answers. In order to find the x component of the applied force I need to know the friction. In order to find the friction I need to find the y component of the applied force, but I can't think of a way to find either.

I thought of trying to solve it by considering energy but without knowing the velocity and given that friction's involved is that even possible?

If I understand correctly Work(Force applied) + Work(Gravity) + Work(Friction) = 0? But how can that lead me to the answer?

 

[PeroK]

I can't see clearly what you have written in that diagram. I'm also not sure what you don't understand.

Suppose the force $F$ was up the slope and there was no friction. Could you answer the question in that case.

Unless the block gains KE, then the net work done on it is zero.

 

[I_Try_Math]

I can't see clearly what you have written in that diagram. I'm also not sure what you don't understand.

Suppose the force $F$ was up the slope and there was no friction. Could you answer the question in that case.

Unless the block gains KE, then the net work done on it is zero.

Yes, if there was no friction, I think I could solve it. Is there a way to solve for the work done by friction using energy considerations so that I don't need to know the components or magnitude of the friction force?

 

[kuruman]

You have the coefficient of kinetic friction $\mu_k.$ What equation do you know that gives you the force of kinetic friction in terms of it? Look it up if you don't remember.

 

[PeroK]

Yes, if there was no friction, I think I could solve it. Is there a way to solve for the work done by friction using energy considerations so that I don't need to know the components or magnitude of the friction force?

You have to resolve the forces into components tangential and normal to the slope. You should not be reluctant to do this.

 

[I_Try_Math]

You have the coefficient of kinetic friction $\mu_k.$ What equation do you know that gives you the force of kinetic friction in terms of it? Look it up if you don't remember.

F_ay = y component of the force applied
w_y = y component of the weight = 340

F_k=0.4*N= 0.4*(F_ay + w_y)

 

[I_Try_Math]

You have to resolve the forces into components tangential and normal to the slope. You should not be reluctant to do this.

I have that the y component of the weight normal to incline is -340 and the x component tangential to the incline is -196.

 

[kuruman]

I have that the y component of the weight normal to incline is -340 and the x component tangential to the incline is -196.

The numbers are correct but they need units. So what is the force of friction?

 

[I_Try_Math]

The numbers are correct but they need units. So what is the force of friction?

F_ay = y component of the force applied
w_y = y component of the weight = -340

F_k = force of kinetic friction

F_k=0.4*N= 0.4*(F_ay + w_y)

I'm stumped on how to find F_ay.

 

[haruspex]

I'm stumped on how to find F_ay.

There are two parts to that, the value of F and the relationship between F and F_ay. Which part are you stuck on?

 

[I_Try_Math]

There are two parts to that, the value of F and the relationship between F and F_ay. Which part are you stuck on?

I can see that F_ay = 340 - N. So then the question becomes how to find N which I cant think how to find.

 

[erobz]

Think "constant velocity" and examime $\nwarrow \nearrow \sum F$.

 

[I_Try_Math]

Think "constant velocity" and examime $\nwarrow \nearrow \sum F$.

Sorry I'm not sure I follow. I see that the net force will be 0 in both the y and x directions if that's what you're getting at.

 

[erobz]

Sorry I'm not sure I follow. I see that the net force will be 0 in both the y and x directions if that's what you're getting at.

Yes, so you should be able to write a system of two equations in two unknowns.

 

[haruspex]

I can see that Fay = 340 - N.

That depends on your sign convention. If you are taking the positive directions so as to make Fay and N both positive then no.
In the y direction, which way does each act, up or down?

 

[I_Try_Math]

Yes, so you should be able to write a system of two equations in two unknowns.

I must be making an algebra error or something. I keep coming up with one equation in two unknowns.

F_ax - 136 + 0.4F_ay = 196

 

[I_Try_Math]

That depends on your sign convention. If you are taking the positive directions so as to make Fay and N both positive then no.
In the y direction, which way does each act, up or down?

The way I tried to have it set up was that N would be positive and F_ay would be negative.

 

[erobz]

I must be making an algebra error or something. I keep coming up with one equation in two unknowns.

F_ax - 136 + 0.4F_ay = 196

Lets work in all variables, sub values at the end. The frictional force is proportional to $N$ (the normal force). $N$ and $F$ should appear in both equations.

Also, please take a moment to learn LaTeX Guide to format your mathematics. It usually doesn't take long to figure out.

 

[I_Try_Math]

Lets work in all variables, sub values at the end. The frictional force is proportional to $N$ (the normal force). $N$ and $F$ should appear in both equations.

Also, please take a moment to learn LaTeX Guide to format your mathematics. It usually doesn't take long to figure out.

Ok so as a starting point I have:

$$N+w_y+F_{ay}=0$$
$$F_{ax}+\mu_k N+w_x=0$$

Does that look correct?

 

[haruspex]

The way I tried to have it set up was that N would be positive and F_ay would be negative.

ok.

 

[haruspex]

Ok so as a starting point I have:

$$N+w_y+F_{ay}=0$$
$$F_{ax}+\mu_k N+w_x=0$$

Does that look correct?

Yes. And in there you know the relationship between the x and y components in each case, so the only unknowns are N and F. Manipulate them to eliminate one and hence find the other.

 

[I_Try_Math]

Yes. And in there you know the relationship between the x and y components in each case, so the only unknowns are N and F. Manipulate them to eliminate one and hence find the other.

$$N(1- \mu_k) + F_{ay}-F_{ax}+w_y-w_x=0$$

I can write N in terms of $F_{ay}$ and $w_y$ and I know the friction coefficient and $w_x$ and $w_y$ so as far as I can tell I have one equation in two unknowns?

 

[erobz]

$$N(1- \mu_k) + F_{ay}-F_{ax}+w_y-w_x=0$$

I can write N in terms of $F_{ay}$ and $w_y$ and I know the friction coefficient and $w_x$ and $w_y$ so as far as I can tell I have one equation in two unknowns?

Go back a step to post 19. Write $F_{ax} , F_{ay}$ in terms of $F$ and the angle of the slope $\theta$.

How many equations, how many unknowns?

EDIT. I think you have sign errors in those equations. All the forces are positive...? What is your positive coordinate system ( i.e which directions are assumed positive)?

 

[I_Try_Math]

Go back a step to post 19. Write $F_{ax} , F_{ay}$ in terms of $F$ and the angle of the slope $\theta$.

How many equations, how many unknowns?

EDIT. I think you have sign errors in those equations. All the forces are positive...? What is your positive coordinate system ( i.e which directions are assumed positive)?

I tried to set it up so that N would be in the positive direction y and downhill would be in the negative x.

 

[erobz]

I tried to set it up so that N would be in the positive direction y and downhill would be in the negative x.

Is this what you think you are using as the coordinate convention?

 

[I_Try_Math]

View attachment 339662

Is this what you think you are using as the coordinate convention?

Yes that's correct. That's what I tried to do at least

 

[erobz]

Yes that's correct. That's what I tried to do at least

yeah, but you are only getting the sign of 1 of the 3 forces correct in each equation...

in the x direction do you think friction and weight point up the hill in positive $x$? Because that is what your equation says.

 

[I_Try_Math]

yeah, but you are only getting the sign of 1 of the 3 forces correct in each equation...

in the x direction do you think friction and weight point up the hill in positive $x$? Because that is what your equation says.

I see the friction in my equation has the wrong sign. I had set $w_{x} =-196$ so I believe that should be correct?

Sigh ...I've redrawn the free body diagram so many times and tried to manipulate the equations I guess wrote them down wrong or did the diagram wrong at some point.

But I can see with your advice of writing the components of F in terms of the magnitude(F) and sin and cos that the system of equations must be possible to solve.

 

[haruspex]

$$N(1- \mu_k) + F_{ay}-F_{ax}+w_y-w_x=0$$

I can write N in terms of $F_{ay}$ and $w_y$ and I know the friction coefficient and $w_x$ and $w_y$ so as far as I can tell I have one equation in two unknowns?

But you did not eliminate N. Write one equation in the form $N=$some expression not involving N, then use that to replace N in the other. This is standard technique for solving simultaneous equations.