# Homework Help: Statics problem: towing force and friction

1. Nov 14, 2015

### jaguar ride

1. The problem statement, all variables and given/known data
Apologies in advance if this problem has been brought here before, I searched and couldn't find it.

The automobile has a mass of 1.7 Mg and center of mass at G. Both the front and rear brakes are locked.
Take μs = 0.3
Determine the towing force F required to move the car.

2. Relevant equations
ΣM = 0
ΣFx = 0
ΣFy = 0
friction = μs * normal force

3. The attempt at a solution

The previous question to this was identical (although with different values), but with only the rear brakes locked. I summed the moments about A to find the reaction at B (normal force), and then multiplied that by the coefficient of static friction to find the frictional force, and solved for F using the horizontal force equation.

I tried the same approach for this problem. Summed the moments about A to find the reaction at B, and summed the forces in the Y-direction to find the reaction at A. I then multiplied those by the coefficient of friction to find their respective frictional forces. Then I summed the forces in the X-direction to solve for F. Wrong.

Clearly this approach is incorrect. Am I way off?

My thoughts are: the frictional force at A is greater than at B, will that have an effect? Does tipping somehow come into play in this problem?

Thanks in advance for any help you can offer. I tried to follow the guidelines, but I can show more work if need be.

2. Nov 14, 2015

### SteamKing

Staff Emeritus
It would be desirable to see your calculations for F.

3. Nov 15, 2015

### jaguar ride

The 16660 N is the 1.7*1000*9.8

4. Nov 15, 2015

### billy_joule

A term is missing from your ∑Ma = 0 expression and your ∑Fx = 0 expression.
Drawing a FBD may help...

5. Nov 15, 2015

### jaguar ride

I do have one, I didn't include it though. I'm failing to see what term I'm missing. I'm using the same technique for the problem I described initially, with the addition of the front wheels' friction. What other force is there to add to my moment equation without giving me too many unkowns? As for the ∑Fx = 0, what other forces are acting in this direction besides both the frictional forces and x-component of F?

Can someone explain why this problem might be different than the other one I mentioned?
(oops, the distance between A and where F is applied is 0.75)

6. Nov 15, 2015

### billy_joule

Sorry, that was a typo, it should have been ∑Fy = 0.

Your FBD is correct.
Your moment equation is missing the moment about A due to 'F' and a similar term is missing from the ∑Fy = 0 equation

7. Nov 15, 2015

### SteamKing

Staff Emeritus
Your original approach assumed that the force applied by the tow rope had no effect on the reactions at the wheels A and B.

Since the force in the tow rope has an upward component, the reactions at A and B in the towing situation do not have the same values as when the car is just sitting there.

8. Nov 15, 2015

### jaguar ride

So if include the force of the tow rope in my moment equation, I won't be able to solve for the reactions will I?

Is determining the frictional forces unimportant in this case? I'm pretty confused now.

9. Nov 15, 2015

### SteamKing

Staff Emeritus
Sure you will.

Just before the car starts to move, it is in static equilibrium. The force F is related to the reactions at A and B by the coefficient of friction and the angle of the towline, so you should be able to write the equations of equilibrium for the car and solve for the value of F.
You don't need to determine the actual friction forces at each wheel, unless you want to. You know the relationship between the friction forces at each wheel and the force in the towline, which is what you are looking for.

You still use ΣF = 0 and ΣM = 0, but you can't ignore the presence of the towline force F when you write these equations.

10. Nov 15, 2015

### jaguar ride

I'm lost now. If I sum the moments about A, then I won't have a reaction at A in my moment equation. Yet I'll still need to know that for the rest of my force equations, won't I?

To top it off, a classmate just told me I need to assume the normal forces are equal, and solve it without even using a moment equation. That can't be right, can it?

11. Nov 15, 2015

### billy_joule

Why not try it and find out?
Your approach in post #3 is correct and will lead to the answer, all you need to do is include the missing terms mentioned. You will find the reactions in terms of 'F', you don't need to find the numerical values of the reactions to find F.

Yeah, that's not right, don't listen to them.

12. Nov 15, 2015

### jaguar ride

Ok, am I on the right track with these equations? I only have two more attempts to answer this so I can't risk it

EDIT: my vertical force equation includes -16660, I forgot to add it until after I'd uploaded the photo.

13. Nov 15, 2015

### SteamKing

Staff Emeritus
I would say you have approximately the correct approach.

However, the points of application of the reactions A and B, the towing force F, and the weight of the car G, are located at different points with respect to one another. I think you moment equation needs to be more general than you have shown here. You should use the vector form of the moment, M = r × F , here when writing the moments of the various forces about a common reference point.

14. Nov 15, 2015

### jaguar ride

I don't understand what you mean by more general. For the moment cause by F, do I not need to use both the components? If I use the cross product rather than what I'm showing here, I'm still ending up with the same answer for that... I'm sorry, It's been a very long day for me and it's hard to understand exactly what you're implying when you say that.

I'm not sure why I'm having such a hard time understanding only this problem. It's making me feel like I've never done statics before.

15. Nov 15, 2015

### SteamKing

Staff Emeritus
M = r × F by definition uses all components of r and F to determine M.

By 'more general', I mean use the cross product to calculate all the moments present in the free body of the car with the tow line attached.

Are you? You can't know that until you make the calculations.

16. Nov 15, 2015

### jaguar ride

Alright, I ended up getting it correct, but this entire problem was really confusing for some reason, and I'm still not sure I actually understand it completely. Thank you both for helping me (or at least trying to). I'm going to talk to my professor in the morning when my brain is actually functioning at a normal level and see if I can really get a grasp on it.

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