Impulse-momentum problem (tractor trailer)

[JJBladester]

Homework Statement

To preface this post, I have already fully solved the problem and obtained the "correct" answer. I am just bothered by the answer after listening to a classmate argue against it.

A trailer truck with a 2000-kg cab and an 8000-kg trailer is traveling on a level road at 90km/h. The brakes on the trailer fail and the antiskid system of the cab provides the largest possible force which will not cause the wheels of the cab to slide. Knowing that the coefficient of static friction is 0.65, determine (a) the shortest time for the rig to come to a stop, (b) the force in the coupling during that time.

Answers:
(a) 19.6s
(b) 10.2 kN

Homework Equations

$$F=ma$$

$$a=\frac{dv}{dt}$$

The Attempt at a Solution

Given:

$$m_c=2000kg$$

$$m_t=8000kg$$

$$v=90\frac{km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=25\frac{m}{s}$$

$$\mu_s=0.65$$

$$f_s=f_{s_{max}}=\mu_smg$$

$$C=coupling\; force$$

Cab:

$$F=ma$$

$$C-f=m_ca$$

$$C-\mu_sm_cg=m_ca$$

$$C-\mu_sm_cg=m_c\frac{dv}{dt}$$

$$Cdt-\mu_sm_cgdt=m_cdv$$

$$\int_{0}^{t}Cdt-\int_{0}^{t}\mu_sm_cgdt=\int_{v_1}^{v_2}m_cdv$$

$$Ct-\mu_smgt=m_cv_2-m_cv_1$$

$$(Eqn 1)\; \; \; \rightarrow \; \; \; \ m_cv_1+Ct-\mu_smgt=m_cv_2$$

Trailer:

$$F=ma$$

$$-C=m_ta$$

$$-C=m_t\frac{dv}{dt}$$

$$-Cdt=m_tdv$$

$$\int_{0}^{t}-Cdt=\int_{v_1}^{v_2}m_tdv$$

$$(Eqn 2)\; \; \; \rightarrow \; \; \; \ m_tv_1-Ct=m_tv_2$$

2 eqns with 2 unknowns. Add Eqn 1 to Eqn 2 and we have:

$$(m_c+m_t)v_1-\mu_smgt=(m_c+m_t)v_2$$

$$t=\frac{(m_c+m_t)(v_1-v_2)}{\mu_sm_cg}=19.6s$$

That is the time it takes for the rig to come to a stop.

The coupling force during that time is:

$$C=\frac{m_t(v_1-v_2)}{t}=10204N=10.2kN$$

_____________________________________________________________________
That is the full "correct" solution. Now, what a classmate argued was that the coupling force C could not be the same in both the cab and the trailer because there is a frictional force on the cab so the forces are not "balanced". He said that we couldn't be assured that the acceleration of both the trailer and the cab were the same.

If you look at the FBD of the cab, you have a coupling force and frictional force acting in the x-direction while the trailer has only a coupling force in the x-direction. Thus, if the coupling forces are equal and opposite (Newton's third law), the acceleration of the trailer must be different than that of the cab.

I may be mangling his reasoning, but it got me thinking "how is C in the cab equal to C in the trailer if there's friction in the x-dir in the cab, but not the trailer?"

 

[tvavanasd]

Cab and trailer are just convenient breaking points for the problem. Why not consider the coupling load at the connection of the license plate to the back of the trailer?
Stop mangling - check your work (subst. the stop time into both eqns 1 and 2). There is a reason why your classmate is not instructing.
What would happen if they accelerated at different rates?

 

[JJBladester]

I understand intuitively that since they are connected by a rigid couple (likely, some kind of metal pin holding the cab to the trailer), that they cannot accelerate differently. However, I am still questioning my original reasoning.

"(subst. the stop time into both eqns 1 and 2)"

What does that get me?

Eqn 1:
$$(2000)(25)+(10204)(19.6)-(0.65)(2000)(9.81)(19.6)=(2000)(0)$$

The left-hand side is equal to 39.6, not 0... So the equality is invalid?

Eqn 2:

$$(8000)(25)-(10204)(19.6)=(8000)(0)$$

Again, what am I supposed to be seeing here?

 

[tvavanasd]

Your significant digits are off - check again.

 

[JJBladester]

Your significant digits are off - check again.

Where are my sig figs off?

The masses of the cab and trailer, the coefficient of friction, the initial velocity, gravity, and friction are all given. So the only two things that could have inaccurate sig figs are the time (t) and the coupling force (C).

Anyway, where are you trying to lead me by plugging (t) back into Eqn's 1 & 2? Should I be getting 0 = 0 and that means something?

 

[tvavanasd]

Where are my sig figs off?

The masses of the cab and trailer, the coefficient of friction, the initial velocity, gravity, and friction are all given. So the only two things that could have inaccurate sig figs are the time (t) and the coupling force (C).

Anyway, where are you trying to lead me by plugging (t) back into Eqn's 1 & 2? Should I be getting 0 = 0 and that means something?

It has been a while for me, but I think the maximum number of signifant figures in the original question is 2 (could be 3). You final answer cannot be more accurate than that.
In your checking, you have used 19.6 seconds as your time (3 sig digs), but 10204 N as the force (5 sig digs), and for gravity you have only used 3 sig digs. Is it practical to expect your answer to be to 5 significant digits? Are you concerned about being within 10 lbf when you have calculated that the force in the coupling is nearly 2300 lbf?

Why would you ever try to make 0=0?
Calculate the coupling force C using both eqn's 1 and 2. I just did it using Excel, and my numbers were dead on. When calculating by hand, it is important to pay attention to significant figures.

I get:
From eqn 1 (cab)
C = (0.65*2000*9.81*19.6-2000*25)/19.6
= 10.2 kN (10201.98 N, but you can't use all digits)

From eqn 2 (trailer)
C = 8000*25/19.6
= 10.2 kN (10204.08 N, but again...)

The point in all of this is that the magnitude of the force in the coupling is the same as seen by the trailer and by the cab.

How long would it take to stop the cab alone? Answer = 3.92 seconds; the cab would decel at 0.65 * 9.81 m/s^2. In your calcs above, if you make the mass of the trailer equal to zero, you should get the same, and the load on the coupling would be zero.

 

[cjl]

Without going through and checking all your work (so I can't comment on its validity right now), here's my response to your statements at the end of the OP:

That is the full "correct" solution. Now, what a classmate argued was that the coupling force C could not be the same in both the cab and the trailer because there is a frictional force on the cab so the forces are not "balanced". He said that we couldn't be assured that the acceleration of both the trailer and the cab were the same.

Yes we can. Since they are coupled, the acceleration of the trailer and cab must be the same, since that is effectively part of the problem definition. It isn't solvable unless you make the assumption that the acceleration must be the same in both the trailer and the cab (which must be true if they stay coupled together). In addition, the coupling force of the trailer on the cab must be the same as the coupling force of the cab on the trailer. This is a simple case of Newton's third law.

If you look at the FBD of the cab, you have a coupling force and frictional force acting in the x-direction while the trailer has only a coupling force in the x-direction. Thus, if the coupling forces are equal and opposite (Newton's third law), the acceleration of the trailer must be different than that of the cab.

Not at all.

If you have 2 masses coupled together rigidly, of masses M₁ and M₂, and you have a force F acting on M₁ only, you can solve for the forces involved (I'll assume everything is in the X direction, and the only forces are the coupling forces {F_c} plus the force applied to mass 1):

$$A_1 = A_2$$

$$F_1*M_1 = F_2*M_2$$

$$F_1 = F - F_c$$

$$F_2 = F_c$$

$$(F - F_c)*M_1 = F_c*M_2$$

$$F*M_1 = F_c*(M_1 + M_2)$$

$$F_c = F*M_1/(M_1 + M_2)$$

Note that if the coupling force and the applied force (friction in this case) follow this relationship, then the acceleration can be the same and the coupling force on both masses can also be the same.

I may be mangling his reasoning, but it got me thinking "how is C in the cab equal to C in the trailer if there's friction in the x-dir in the cab, but not the trailer?"

Because of Newton's third law.

The friction on the cab just tells you that the system as a whole is accelerating, but there's no reason why it would break symmetry in the coupling.

 

[cjl]

OK, I've gone through all your work now, and your overall solution looks good, if a bit overcomplicated. Also, the point tvavanasd is trying to get at is that to the precision you have, your equations check out. Yes, when you plug into equation 1, you get 39.6 = 0, but when you figure out your significant digits properly, you quickly discover that to the precision you have, 39.6 is equal to zero.

Specifically, you have (2000)(25)+(10204)(19.6)-(0.65)(2000)(9.81)(19.6)=(2000)(0). If you work through and evaluate this, even if you are generous with the sig. figs, you get the following:

5.0*10⁴ + 2.00*10⁵ - 2.5*10⁵ = 0.

Note that when this is evaluated, you get precisely 0 = 0. You were carrying through precision that you never had, which is why you got 39.6.

 

[JJBladester]

From eqn 1 (cab)
C = (0.65*2000*9.81*19.6-2000*25)/19.6
= 10.2 kN (10201.98 N, but you can't use all digits)

From eqn 2 (trailer)
C = 8000*25/19.6
= 10.2 kN (10204.08 N, but again...)

The point in all of this is that the magnitude of the force in the coupling is the same as seen by the trailer and by the cab.

Ok, so it does make sense to re-evaluate Eqn 1 and Eqn 2 to find that C is the same in both cases, after we have solved for time (t).

How long would it take to stop the cab alone? Answer = 3.92 seconds; the cab would decel at 0.65 * 9.81 m/s^2. In your calcs above, if you make the mass of the trailer equal to zero, you should get the same, and the load on the coupling would be zero.

Ok, I understand removing the trailer from Eqn 1 and Eqn 2, then setting them equal and solving for time (t). I have, then, t=v₁/(μ_sg)=3.92s as you do above, except that you may have mis-wrote your equation for the decel of the cab... But I get your point. Thank you!

 

[JJBladester]

Note that when this is evaluated, you get precisely 0 = 0. You were carrying through precision that you never had, which is why you got 39.6.

Ok, so is there some kind of "rule-of-thumb" that when you get 0=0 (obviously a true equality), you know your equation is consistent? I think tvavanasd was saying I shouldn't be worried with that?

Why would you ever try to make 0=0?

 

[cjl]

Ok, so is there some kind of "rule-of-thumb" that when you get 0=0 (obviously a true equality), you know your equation is consistent? I think tvavanasd was saying I shouldn't be worried with that?

The general rule of thumb is that if you started with a certain set of equations at the beginning of the problem, and you ended up with a certain set of solutions at the end, you should be able to plug in your set of solutions into your original equations and get a true equality (which doesn't need to be 0 = 0, it could just as easily be 2873/32 = 2873/32). If you plug your solutions into your original equations and you don't get an equality, then something has gone wrong.

 

[cjl]

...except that you may have mis-wrote your equation for the decel of the cab... But I get your point. Thank you!

What makes you say that? The deceleration of the cab would simply be 0.65*g in that case. This is pretty easily shown.

$$F_n = W = mg$$

$$F_f = \mu*F_n$$

$$a = F/m$$

$$a = \mu*m*g/m = \mu*g$$

(Note that this is only valid if the entire mass of the system is acting on the surfaces with friction - in the context of this problem, it only works if the mass of the trailer is zero)

 

[JJBladester]

The general rule of thumb is that if you started with a certain set of equations at the beginning of the problem, and you ended up with a certain set of solutions at the end, you should be able to plug in your set of solutions into your original equations and get a true equality (which doesn't need to be 0 = 0, it could just as easily be 2873/32 = 2873/32). If you plug your solutions into your original equations and you don't get an equality, then something has gone wrong.

This is good advice.