2 Exercises with Forces -- dogs pulling a sled

[albertus50]

Homework Statement

2 dogs are pulling a sled at a constant speed. The movement is happening on a horizontal flat surface. The Sled weighs 150kg. The cinetic frictional coefficient on the snow is 0.05.
Determine the tension force on the leash of each dog.

Relevant Equations

The teacher has not given us any exact equations we need to use but i know we used f = mu N. However, any formula used is fine, i am ready to understand the formula as long as the answer is right.

Hello ! we have recently been given the exercice stated above, along with a diagram that i inserted as a picture.The problem i have with this exercice is that i am not quite sure whether my idea about how to do it was right. I was thinking about replacing the normal force (N) in the F= mu x N formula by the gravitational force of the sled because from what i understood it is equal to the gravitational force except it points up. Except, now that i know the friction force i do not know how to proceed because i can not Newton's second law as i don't know the acceleration, and thus i do not know how to find the force that the 2 dogs are using to pull the sled. Thank you for reading and thank you in advance for your help!

 

[TSny]

i can not Newton's second law as i don't know the acceleration

Information about the acceleration is contained in the statement:

2 dogs are pulling a sled at a constant speed.

 

[albertus50]

Allow me to correct myself: there is no acceleration so I can not use F=MA if I could I would just substract the friction force but without that I have no idea how to proceed

 

[TSny]

Allow me to correct myself: there is no acceleration so I can not use F=MA if I could I would just substract the friction force but without that I have no idea how to proceed

F = MA is still applicable when the acceleration is zero.

This is a problem where vectors are involved. The second law in vector form is $\vec F_{\rm net} = m \vec a$. This is often written as $\sum \vec F = m \vec a$. If you introduce a Cartesian coordinate system, you can write the vector equation as three separate component equations: $$\sum F_x = m a_x$$ $$\sum F_y = m a_y$$ $$\sum F_z = m a_z$$

 

[TSny]

I recommend applying the second law twice: once for the sled and once for the knot where the ropes come together.

Begin by drawing free-body diagrams for the sled and knot.

 

[albertus50]

Sorry for my bad drawing but i represented it first from a side pov and then as viewed from above. The problem is i do not understand how to apply it to each component individually Because the acceleration of each is still 0 and that would lead to F=MA. If i divide in 3 with Fx, Fy and Fz then the only acceleration i have is 9.81m/s^2 but i am looking for the one in Fx... sorry i am kind of slow with this type of stuff

 

[scottdave]

It's the vector sum of all forces = mass x acceleration. So which direction does your frictional force point? As viewed from the sled, which direction does the rope tension point?

 

[TSny]

OK. Your diagram for the sled alone looks good. Later you will want to "zero in" on the knot and draw a diagram just for the knot.

The next step is to add a coordinate system to the sled diagram. Suppose you choose the z-axis to point upward, perpendicular to the ground and the x-axis as horizontal in the direction of your tension F_t.

Try to apply the equation $\sum F_z = ma_z$ to the sled and use your diagram to see what forces go on the left side.

 

[albertus50]

It's the vector sum of all forces = mass x acceleration. So which direction does your frictional force point? As viewed from the sled, which direction does the rope tension point?

My frictional force points in the opposite ditection of the rope tension so that would mean that the only resulting force is my tension force except it is a bit diminished because of the friction right?

 

[albertus50]

OK. Your diagram for the sled alone looks good. Later you will want to "zero in" on the knot and draw a diagram just for the knot.

The next step is to add a coordinate system to the sled diagram. Suppose you choose the z-axis to point upward, perpendicular to the ground and the x-axis as horizontal in the direction of your tension F_t.

Try to apply the equation $\sum F_z = ma_z$ to the sled and use your diagram to see what forces go on the left side.

So on the z axis, the only force that exists is my gravitational force, so that would give us Fz=M x Az which is Fz = 150 x 9.81 = 1471.5N for my force on the Z axis right?

 

[scottdave]

So on the z axis, the only force that exists is my gravitational force, so that would give us Fz=M x Az which is Fz = 150 x 9.81 = 1471.5N for my force on the Z axis right?

If there is only 1 force in a direction (or a net force), then it will accelerate in that direction. So what keeps it from sinking into the snow/ice? Hint: you already talked about it in a previous post.

As to your question (post #9 above) about "slightly diminished" force, don't confuse speed and acceleration. How much acceleration is "constant speed" ?

 

[albertus50]

Right, i think what you're referring to is the normal force? But the normal force compensates for the gravitational force so does that mean my sum of forces on the Z axis is equal to 0?

As to your question (post #9 above) about "slightly diminished" force, don't confuse speed and acceleration. How much acceleration is "constant speed" ?

About this, if i am going at constant speed my acceleration is 0 because otherwise the speed would keep increasing right?

 

[jbriggs444]

About this, if i am going at constant speed my acceleration is 0 because otherwise the speed would keep increasing right?

Right.

Ignoring the possibility that acceleration is negative so that your speed will decrease and then increase as you begin moving backward. In which case speed would eventually keep increasing anyway.

And ignoring the possibility that direction is changing so that acceleration could be non-zero in spite of constant speed. But we can assume straight line motion for this problem.

 

[albertus50]

So in this case it is as i doubted, i can not use the 2nd law because my acceleration is 0 in every direction except y (perpendicular to the ground), and the F that i find if the law is applied to that direction is not the one i am looking for since i need the F horizontally (x axis). That's why i am kinda stuck here

 

[jbriggs444]

So in this case it is as i doubted, i can not use the 2nd law because my acceleration is 0 in every direction except y (perpendicular to the ground), and the F that i find if the law is applied to that direction is not the one i am looking for since i need the F horizontally (x axis). That's why i am kinda stuck here

Newton's second law applies perfectly well when acceleration is zero. Zero is a number.

What makes you say that your vertical acceleration is non-zero?

 

[albertus50]

If i use my acceleration as 0 though my force will be 0 horizontally and i thought the objective was to find my force horizontally to find the tension force in the cable.

i thought my vertical acceleration is non zero because there is a force of gravitation puling me down and i assumed that thus my acceleration in the F=MA equation could not be zero but i now realize it was wrong... so i have no acceleration in any direction and I do not know of a formula that could allow me to find the tension force without knowing acceleration / speed etc...

 

[jbriggs444]

If i use my acceleration as 0 though my force will be 0 horizontally and i thought the objective was to find my force horizontally to find the tension force in the cable.

Exactly which force on the drawing is "my force"?

 

[albertus50]

Sorry about that, i meant the force i am looking for so the tension force in the cable. i was hoping to get that by finding the sum of forces with Newton's second law and then taking the friction force out of the sum

 

[jbriggs444]

Sorry about that, i meant the force i am looking for so the tension force in the cable. i was hoping to get that by finding the sum of forces with Newton's second law and then taking the friction force out of the sum

So let's do that. Find the sum of the forces on the sled with Newton's second law. What must the sum be?

 

[albertus50]

So my friction force is 73.5, if i use the equation f= mu x N, mu being 0.05 and N 1471.5N (9.81 x 150kg). So my sum of forces would be Ft - 73.5, Ft being my unknown tension force, and if i try to put that into Newton's second law equation it gives me
Ft - 73.5 = 150 x 0 so Ft = 73.5 but that would mean that my tension force is equal to friction so it doesn't really work. I thought about using 9.81 as my acceleration but that would not really make sense to me, and as long as my sled is going at constant speed one side of my equation will be equal to 0

 

[jbriggs444]

but that would mean that my tension force is equal to friction so it doesn't really work.

Why not? That is a correct computation for the tension force acting forward on the sled.

But you were not asked for the tension force acting forward on the sled, were you? See post#5 by @TSny

 

[albertus50]

if my tension force is equal doesn't that mean that together they make 0 and thus there is no force acting upon the sled?
Is there any way for me to find the forces of tension on the knot without finding the force of tension in the cable pulling the sled?

 

[jbriggs444]

if my tension force is equal doesn't that mean that together they make 0 and thus there is no force acting upon the sled?

Yes. The force on the sled is zero. Just as Newton's law says it should be. The sled is not accelerating. The net force on it must be zero.

You solved the equations on the assumption that it was zero. It should be zero. You should not be shocked to discover that it is zero.

Is there any way for me to find the forces of tension on the knot without finding the force of tension in the cable pulling the sled?

You calculated the tension in the cable pulling the sled already. It is not zero.

 

[albertus50]

Oh my god i finally get it thank you so much, so the tension in the cable is 73.5? and then in order to find the tension in each one of the cables should i use trigonometry and do cos(30)=73.5/x, x being my tension in the cable linked to the leash?

 

[jbriggs444]

Oh my god i finally get it thank you so much, so the tension in the cable is 73.5? and then in order to find the tension in each one of the cables should i use trigonometry and do cos(30)=73.5/x, x being my tension in the cable linked to the leash?

Bearing in mind that there are two dogs, each with a cable.

See post #6 by @TSny. This is Newton's second law applied to the knot. The $\cos 30$ comes in because you are considering the fore-and-aft component only. If you were considering the side-to-side component you'd be using $\sin 30$.

 

[albertus50]

Bearing in mind that there are two dogs, each with a cable.

Does that mean i should divide by two?

 

[jbriggs444]

Does that mean i should divide by two?

Yup!

 

[albertus50]

So that means i need to use (cos(30)=73.5/x)/2 and that's the right formula? Thanks a lot for explaining !

 

[jbriggs444]

So that means i need to use (cos(30)=73.5/x)/2 and that's the right formula? Thanks a lot for explaining !

Yes.

I would have written it differently, adhering to the standard form of Newton's second law. And I'd have kept things in algebraic form until ready to evaluate the result. And a little commentary never hurts. Don't make the reader guess at what you are doing. Explain it.

Start with Newton's second law. $$\vec{F} = m \vec{a}$$Apply it to the knot in the rope. There are three forces at the knot in the rope. We can call these $T$ from the sled, $t_1$ from dog 1 and $t_2$ from dog 2. The mass of the knot is zero and its acceleration is zero. Substituting into the previous equation, this yields:$$\vec{T} + \vec{t_1} + \vec{t_2} = 0$$This is a vector equation. All three components of the vector on the left hand side must be equal to the corresponding component on the right hand side. We could choose any component to proceed. But it seems most useful to pick a component where our known value $\vec{T}$ is non-zero. Otherwise that information would disappear from our equation. So let us pick the fore and aft direction and call it "$x$".$$T_x + t_\text{1x} + t_\text{2x}=0$$The x components of the two dog tensions both point rightward and are both equal to $\cos \theta$ times the magnitude of the full dog tension. The x component of the sled tension is the full magnitude of the sled tension. We've been a bit careless with sign conventions up to now. Let's take the opportunity to fix that up. Tensions are always positive scalars. The associated forces take their vector values from the direction the rope is facing at the attachment point. We can use a standard right=positive convention.$$−T+t_1 \cos \theta + t_2 \cos −\theta=0$$I took a little care use a negative angle since dog 2 is below the x axis. Of course, that unnecessary since the cosine is an "even" function -- it has gives the same result for negative inputs as for the corresponding positive inputs. So $\cos \theta=\cos −\theta$. We can combine the two dog tension terms together and use $t_d$ to refer to the tension in each dog rope.$$−T + 2 t_d \cos \theta=0$$We are solving for $t_d$ so the goal is to write an equation with $t_d$ alone on the left. So we add $T$ to both sides and then divide both sides by $2 \cos \theta$. That yields$$t_d=\frac{T}{2 \cos \theta}$$That equation is good as far as it goes. But we had previously used Newton's second law to arrive at the sled tension $T$. That result should have been written as$$T=\mu_k mg$$Where $\mu_k$ is the coefficient of kinetic friction for the sled, $m$ is the mass of the sled and $g$ is the acceleration of gravity. Substituting this value for T in the previous equation, we would get$$t_d = \frac{\mu_k m g}{2 \cos \theta}$$That is the format you should strive for: Your desired unknown alone on the left hand side and a formula with the knowns from the problem statement on the right. Only once you get to this should you start plugging in numbers. Advantages of saving the numbers for last include a reduced chance for errors transcribing numbers and an increased chance of being able to cancel things in the algebra. If you discover that you screwed up [like I just did, dropping a factor of 2 in the last couple of equations] you can go back and fix things without re-doing all of the numbers.

[I like to define each variable name that is used. That is because I am a computer programmer by trade. We were taught to 1. Declare all variables and 2. Document all variables. Lost all my Tex making this final edit. Repaired. Hope I didn't miss anything]

 

[albertus50]

I really can not that you enough for this perfectly clear explanation! I finally understand it now, many thanks for writing it in such a clean and detailed paragraph, it allowed me to see exactly in which places i was getting stuck. I very much appreciate it. Have a great day and thank you again!