Solve Simple Work Problem: Steve + 30° + 0.20 μ

[PurelyPhysical]

Homework Statement

A 50 KG box was dragged across a floor for a distance of 20 meters at a constant velocity by Steve pushing on the box at an angle of 30 degrees below the horizontal. If the coefficient of kinetic friction between the box and the floor is 0.20, calculate the work he did.

Homework Equations

How can you solve for velocity in this situation? Is it necessary? Also, what is the answer to this problem?

The Attempt at a Solution

So far I have set up Delta K + Delta E (thermal) = Work, but I cannot progress because K = 1/2mv^2 has velocity in it, which I cannot find.

 

[Simon Bridge]

Welcome to PF.
Hint: you are given a force and a distance...

 

[PurelyPhysical]

Welcome to PF.
Hint: you are given a force and a distance...

Thank you :)
Am I given a force? I am failing to see where I am given a force. I see a weight, a distance, an angle, and a coefficient of friction. If I had a force I could use Fdcosθ=W, which would sure make things easier, but I can't see how to figure out what the force is.

Edit: I think I see it now. Are you referring to the frictional force? I'm currently trying to work it out.

 

[PeroK]

Homework Statement

A 50 KG box was dragged across a floor for a distance of 20 meters at a constant velocity by Steve pushing on the box at an angle of 30 degrees below the horizontal. If the coefficient of kinetic friction between the box and the floor is 0.20, calculate the work he did.

Homework Equations

How can you solve for velocity in this situation? Is it necessary? Also, what is the answer to this problem?

The Attempt at a Solution

So far I have set up Delta K + Delta E (thermal) = Work, but I cannot progress because K = 1/2mv^2 has velocity in it, which I cannot find.

There are a few "conventions" with a problem like this that you seem to be missing.

a) Gravity is not always mentioned explicitly: it's often assumed to be there. This box is not being pushed across the floor of a space station where there is zero gravity. You may assume normal Earth gravity.

b) "At constant velocity" often indicates that you are not to worry about an acceleration phase at the beginning to get the box moving. You can furthermore assume that the Kinetic Energy of the box is constant throughout, so you needn't worry about KE either.

 

[PurelyPhysical]

There are a few "conventions" with a problem like this that you seem to be missing.

a) Gravity is not always mentioned explicitly: it's often assumed to be there. This box is not being pushed across the floor of a space station where there is zero gravity. You may assume normal Earth gravity.

b) "At constant velocity" often indicates that you are not to worry about an acceleration phase at the beginning to get the box moving. You can furthermore assume that the Kinetic Energy of the box is constant throughout, so you needn't worry about KE either.

Thank you for responding.

So that would leave me to only worry about the force of kinetic friction, μmg(displacement)? The result I come up with is 3675 Joules. I am still getting a feel for what an appropriate value in joules should look like, given various values of force and energy.

My calculations exactly were (.20)(50)(9.8)(20)= 1960 J

 

[PeroK]

Thank you for responding.

So that would leave me to only worry about the force of kinetic friction, μmg(displacement)? The result I come up with is 3675 Joules. I am still getting a feel for what an appropriate value in joules should look like.

First, to be precise, the question gives a "distance" not a "displacement". This is important because the work depends on distance. If, for example, he pushed the box backwards and forwards 10m, then the dispalcement would be 0, but he'd still have done the same work as pushing it forwards 20m.

Where did 3675 come from?

 

[PurelyPhysical]

First, to be precise, the question gives a "distance" not a "displacement". This is important because the work depends on distance. If, for example, he pushed the box backwards and forwards 10m, then the dispalcement would be 0, but he'd still have done the same work as pushing it forwards 20m.

Where did 3675 come from?

Ah, I see. Thank you, that makes sense.

I was looking at the wrong problem, my answer was (.20)(50)(9.8)(20) = 1960 J

 

[PeroK]

I was looking at the wrong problem, my answer was (.20)(50)(9.8)(20) = 1960 J

Where did 1960 come from?

 

[PurelyPhysical]

Where did 1960 come from?

I multiplied the force of kinetic friction by the distance, at least I think I did.

 

[PeroK]

I multiplied the force of kinetic friction by the distance, at least I think I did.

What about the "at an angle of 30 degrees"?

 

[PurelyPhysical]

What about the "at an angle of 30 degrees"?

The force from Steve is at a 30 degree angle. Would that make the frictional force subject to the same angle? If I multiply the cosine of 30 degrees to my previous answer, I get 1697.36

 

[PeroK]

The force from Steve is at a 30 degree angle. Would that make the frictional force subject to the same angle? If I multiply the cosine of 30 degrees to my previous answer, I get 1697.36

This problem may be trickier than you have noticed. Steve is pushing the box down thus increasing the friction. That's why it's important to a) draw a diagram of the situation and b) post more than just a "final answer" of xyz Joules.

You need to think more carefully about the forces involved.

 

[PurelyPhysical]

This problem may be trickier than you have noticed. Steve is pushing the box down thus increasing the friction. That's why it's important to a) draw a diagram of the situation and b) post more than just a "final answer" of xyz Joules.

You need to think more carefully about the forces involved.

I have drawn a diagram. So I need to calculate just how much more kinetic friction there is in addition to the .20 that already exists between the box and ground? You had mentioned that I don't need to worry about Kinetic Energy. I am at a bit of loss as to what to do next, but I am getting a better idea. The work that I should come up with suggests something about what the actual coefficient of friction is, given the additional downard force. Is that correct?

 

[PeroK]

I have drawn a diagram. So I need to calculate just how much more kinetic friction there is in addition to the .20 that already exists between the box and ground? You had mentioned that I don't need to worry about Kinetic Energy. I am at a bit of loss as to what to do next, but I am getting a better idea. The work that I should come up with suggests something about what the actual coefficient of friction is, given the additional downard force. Is that correct?

First, note that "constant velocity" implies balanced forces. That's going to be important.

If he pushed the box horizontally, it would be simple. The work would be simply $\mu mgd$. And, in fact, the force he is applying would be the same as the friction $F = \mu mg$

But, as he is pushing the box down, you will have to set up some equations involving the forces involved. Let $F$ be the force with which he is pushing the box. You now need to try to solve for $F$. To get you started: you need to resolve $F$ into its horizonal and vertical components.

 

[PurelyPhysical]

First, note that "constant velocity" implies balanced forces. That's going to be important.

If he pushed the box horizontally, it would be simple. The work would be simply $\mu mgd$. And, in fact, the force he is applying would be the same as the friction $F = \mu mg$

But, as he is pushing the box down, you will have to set up some equations involving the forces involved. Let $F$ be the force with which he is pushing the box. You now need to try to solve for $F$. To get you started: you need to resolve $F$ into its horizonal and vertical components.

Okay, I am starting to get a better idea of what's going on. I have come up with μmg = 147. I know how to resolve a vector force into it's components, but I am confused about one thing. Would 147 be the x component or would it be the hypotenuse? If it is the x component, how do I go about solving for the rest of the triangle? Have I already made an error?

Am I going to figure out the additional coefficient of friction by using the vertical component of steve's force?

 

[PeroK]

Okay, I am starting to get a better idea of what's going on at this point. I have come up with μmg = 147. I know how to resolve a vector force into it's components, but I am confused about one thing. Would 147 be the x component or would it be the hypotenuse? If it is the x component, how do I go about solving for the rest of the triangle? Have I already made an error?

Am I going to figure out the additional coefficient of friction by using the vertical component of steve's force?

First, it's important to see that Steve is making it harder by pushing down at an angle. This is for two reasons: a) not all of his force is horizontal so not all of it is moving the box and b) the downward force is increasing the frictional resistance.

So, you need to resolve Steve's force $F$ into $F_h$ (which pushes the box forward) and $F_v$ which pushes the box down. So, now you have two forces pushing down on the box: $g$ and $F_v$

One more thing. As problems get harder, it is better to hold off plugging in the numbers. I would leave things like $\mu mg$ for the time being. In any case, I've no idea where 147 came from!

Can you find an expression for the frictional force? Let's call it $F_f$. It will depends on $\mu, m, g$ and $F_v$

 

[PurelyPhysical]

Haha, I keep referencing a similar problem by mistake that only varies by number. My answer for μmg should have been 98, which I hope is right!

This is the expression I've come up with F_f = μmg + μF_v

 

[PeroK]

Haha, I keep referencing the wrong problem. My answer for μmg should have been 98, which I hope is right!

This is the expression I've come up with F_f = μmg + μF_v

Yes. The obvious difficulty now is that you need to find $F_v$. Remember I said that balanced force would be important. If the box is moving at constant velocity, then which forces are balanced?

 

[PurelyPhysical]

Yes. The obvious difficulty now is that you need to find $F_v$. Remember I said that balanced force would be important. If the box is moving at constant velocity, then which forces are balanced?

Now we're getting somewhere!

Steven's force in the horizontal direction and the opposing frictional force?

 

[PeroK]

Now we're getting somewhere!

Steven's force in the horizontal direction and the opposing frictional force?

Yes. Can you see how to solve for $F$ using the relation between $F_v$, $F_h$ and $\theta$?

 

[PurelyPhysical]

Yes. Can you see how to solve for $F$ using the relation between $F_v$, $F_h$ and $\theta$?

I can see that F_v is the y component, and that F_h is the x component. I can see that F_v is 50 percent of F, and F_h is .866 percent of F. I feel like there is something simple that I am failing to see here. I think it involves using tangent, but I am struggling to see how to come up with actual values to put in.

 

[PeroK]

I can see that F_v is the y component, and that F_h is the x component. I can see that F_v is 50 percent of F, and F_h is .866 percent of F. I feel like there is something simple that I am failing to see here. I think it involves using tangent, but I am struggling to see how to come up with actual values to put in.

You have $F_f = \mu(mg + Fsin\theta) = F_h = Fcos\theta$

Is your algebra up to rearranging that into $F = \dots$?

 

[PurelyPhysical]

You have $F_f = \mu(mg + Fsin\theta) = F_h = Fcos\theta$

Is your algebra up to rearranging that into $F = \dots$?

I sure hope so!
F = [μ(mg+sinθ)]/cosθ

I'm still double checking my answer, but I've come up with F = 113.279

 

[PeroK]

I sure hope so!
F = [μ(mg+sinθ)]/cosθ

Not quite. There are two $F$'s in the first equation.

I'm going offline now. Maybe someone else can help if you can't finish it off.

 

[PurelyPhysical]

Not quite. There are two $F$'s in the first equation.

I'm going offline now. Maybe someone else can help if you can't finish it off.

Thank you for the help!

 

[PeroK]

Thank you for the help!

I think this problem may be a step up from what you've done so far. The approach to solving it is quite common and comes up in a lot of problems, so it's worth understanding. Especially the idea that you have to be prepared to think several steps ahead to get the answer: the answer doesn't just come from one equation.

 

[PurelyPhysical]

I think this problem may be a step up from what you've done so far. The approach to solving it is quite common and comes up in a lot of problems, so it's worth understanding. Especially the idea that you have to be prepared to think several steps ahead to get the answer: the answer doesn't just come from one equation.

I'm beginning to see that. I also see that second F I missed, which does make the algebra a bit more tricky, but I think I'll manage. So once I get the force, I just use W=Fdcosθ and the problem is solved?

I've read up to this difficulty in the book and beyond, but I have not actually tried the problems for myself yet. I understand the concepts, but am not as familiar with the math of it all just yet. Thanks again for your help!