We're going to go ahead and use all the tools

we've learned so far to solve a real world problem.

This is the problem I actually introduced to you at the beginning of the course.

You may have found yourself at one time in this unfortunate

situation and so we have a tow truck, weighs 5,000 pounds.

We've show its mass, center at G.

And we're told that it's

rear-wheel driven, so we only have friction on the rear tires.

And the coefficient of friction is 0.25.

And we're asked to find the maximum force we can

develop in the tow cable, to see if we can pull us out of this unfortunate situation.

So here's the free body diagram. I show my tension in my tow cable.

I have my normal forces at both the front tires

and the back tires, that are not necessarily the same.

I've got my weight, which is 5,000 pounds.

There's no front, friction in

the front tire, and so just in the rear tires.

I've got a friction force opposing the impending motion.

and so that's a good free body diagram of

the situation and so where do you go from there?

Okay, what you should do is you should

go ahead and apply the equation of equilibrium.

I'll choose to go ahead and apply the sum of the moments about point B.

I've chosen counter-clockwise positive.

I'd like you to go ahead and assemble that equation.

Okay, now that you've assembled the equation, I've got 5000 times its moment

arm, which is 6, minus N1 times its moment arm, which is 12 and its minus.

Because it has a tendency to cause a clock-wise

rotation, which is opposite the sine convention I have here.

and then what I do is I break the the tension force in the cable into two parts.

it's y component and it's x component and so for its x component it's going

to be 1 over the square root of 2 times T times it's moment arm which is 3.

And then also T over the square root of 2 is the y component.

And it's moment arm is 14.

And so we end up with one equation and two unknowns.

So at that point the question is

what should we do next?

And often the best thing to do is we need more information,

so let's go ahead and do another free another equation of equilibrium.

And so go ahead and at this time and lets

do the sum of the forces in the x direction.

so you've got the x component of the tow cable

which is minus T over the square root of 2.

And we have the f force, the friction force, and that's it in the x direction.

So we now have two equations, but I have one, two, three unknowns.

I've got N1, the tension and f.

And so I still need to get rid of one

of the unknowns, and so my question to you is,

is there anything else we know about the problem

that can help us reduce one of these unknowns?

at, at this rear tire and so the gain of max tension, we'll use F-max which

is equal to the coefficient of friction times

the normal force at that location which is N1.

And so let's

go ahead and substitute that in, we've got u sub s times n

which is the friction force is T over the square root of 2.

And so I end up with N1 equals 4T over the square root of 2.

however, now I have yeah I now I have a equations I can go ahead and solve.

I can substitute 2 into 1, because I have two equations

two unknowns.

I want to solve for T, and so we'll go

ahead and, and, and substitute N1 into our first equation.

And we arrive at T equals 1370 pounds.

So that's the, the maximum tension I can develop in the cable before

the tow truck will start sliding to the right or excuse me, to the left.

And so,

my question to you, is that is that are we

good to go or should we check anything else out.

And hopefully what you thought about was

in friction cases, we can have impending slipping.

Or we can have impending tipping.

So we better check to make sure that this truck hasn't tipped.

And so my question to you is, how will

we be able to tell from this free body diagram

if the truck has indeed tipped.

So what you should have thought about or deduced is that N2 here

is a positive value if the front tires are pushing down on the ground.

However if N2 ends up equalling 0 or a negative

number, that means the front tires have lifted off the ground.

And the truck has tipped and so lets double check to make sure it won't

tip, by summing moments about point 8 and,

and, and, and finding well actually we don't

even need to sum moments about point 8. What we can do is we can say okay

what, what, what was N1 and then we will just sum forces in the y direction.

So N1 was given here we said that the intention that we, we

could develop was 1370 pounds so that means that N1 ends up being

13, 380 300, 3870 pounds. if I sum

forces in the y direction now I find out that N2 will need

to be 2,100 pounds. And so since N2 in this situation ends up

being greater than 0, you find out that yes, indeed the truck doesn't tip.

And as I mentioned earlier, if you didn't want to go through this type

of analysis, you could have also gone

through an analysis where you sum moments about

a as the other equation of equilibrium.

you'd have your tension force in the cable, you'd have your weight,

you could solve for N2 again it would amount to be 2,100 pounds.

And since it's positive you know that the front wheel

is staying on the ground and the truck doesn't tip.

And so the amount of tension that you can develop in the cable, the

max tension that you can able to develop in the cable is 1370 pounds.

maybe as an exercise on your own, think about your

vehicle if you have one or somebody's vehicle and find

out if that's enough to, do you think that would

be enough to pull that, that car out of this situation.