What are the forces on the hitch between truck and trailer?

[samdiah]

Tractor pulling 2 trailers starts from rest and then accelerated to a speed of 16.2 km/h in 15s on a straight, level section of highyway. Mass of truck is 5450 kg, trailer a is 31500 kg and trailer b is 19600 kg. Coefficient of kenetic friction for tractor and trailor is 0.23.

a)What force does truck provide to accelerate the whole system?

b) What are the forces on the hitch between truck and trailer?

ans-see attachment

I need clarification on how to solve

 

[OlderDan]

Show us what you have done.

 

[samdiah]

The work is in the attachment

 

[samdiah]

I am pasting the work but the attachment is better.

Tractor pulling 2 trailers starts from rest and then accelerated to a speed of 16.2 km/h in 15s on a straight, level section of highway. Mass of truck is 5450 kg, trailer A s 31500 kg and trailer Bis 19600 kg. Coefficient of kinetic friction for tractor and trailer is 0.23.

a) What force does truck provide to accelerate the whole system?
.u=0.23
a=v2-v1/t
=16.2/15
=1.08m/s2 acceleration for whole system

FNTRUCK =53 410 N

FNA =308 700 N

FNB =192 080 N

FKTRUCK =0.23*53410 N
= 12 284.3 N

Fnet=ma
=1.08*5450
=5886 N

FAPP TRUCK =12284.3+5886
= 18170.3 N

b) What are the forces on the hitch between truck and trailer?

*Not sure what to do

 

[samdiah]

1. Basically I found the overall acceleration,
2. then found the normal force by assuming normal=force of gravity=mg,
3. then using coefficient of kinetic and normal force figure out the kenetic friction
4. Used f=ma to find out net force.
5. added the friction and netforce to figure out truck's force

 

[OlderDan]

In part a) you found the acceleration. Look at the last trailer and find the force needed to produce that acceleration, then examine the forces acting on that trailer. Then do the same thing for the first traler.

 

[samdiah]

So is part a right or on the right path?

 

[OlderDan]

So is part a right or on the right path?

It looks partially on the right path, but you need to be a lot more careful with your units. You also need to be sure you have included all of the forces acting on each object. It doesn't look like you have.

Draw a free body diagram for each of the 3 parts of the system. Identify all of the forces acting on each part. This will include the unknown hitch forces, as well as the force you are looking for in part a). Apply Newton II to each object and you will get three equations involving the three unknown forces. You can add these equations to do part a), and then work with individual equations to do part b)

 

[samdiah]

"you will get three equations involving the three unknown forces."

I tried to find the equations and got FT3-FfB=mBa
FT2-FfA = mAa
FT1-FfTK= mTKa
But we already have the acceleraton of the system 0.3m/s^2(thanks for the unit reminder) and we also have the friction at each point:
Ff=uk*mg

I don't know how I will solve for anything forthe whole system when I already have numerical answer for each tension in each equation.

I cannot use the simultaneous equations to solvethis because it makes no sense.

Is the force exerted by truck sum of all tensions or just the tension on the truck itself?

 

[OlderDan]

Your equations are missing terms. Every FT has to appear in two equations.

 

[samdiah]

O ok that explains somethings. does every friction have to appear in two equations too then?

 

[OlderDan]

O ok that explains somethings. does every friction have to appear in two equations too then?

No. The point of drawing free body diagrams is to isolate each object and consider only the forces acting directly on that object. Only one friction acts directly on each object. If you change anyone of the three, it affects the motion of all three objects because it results in changing the tensions.

 

[samdiah]

Great!
Now I got the three equations:
FT3 - FfB = mBa
FT3 + FT2 - FfA = mAa
FT3 + FT2 + FT1 - FfTK = mTKa

What I did now is started with B and found the friction on the trailer and plugged it into eq.1 and got a value for tension 3

Did the same to find out the friction of A plugged in value of T 2 did the same thing for truck.

Overall I have three values of tension now. The truck provides the tension of the value of T1.

Are my procedures correct?

 

[samdiah]

Help Please! I know I ask for too much, but wat can I do I am too dumb!

 

[samdiah]

Why do I get a nagative answer? Are my equations wrong is it minus the tension or plus?

 

[OlderDan]

Great!
Now I got the three equations:
FT3 - FfB = mBa
FT3 + FT2 - FfA = mAa
FT3 + FT2 + FT1 - FfTK = mTKa

What I did now is started with B and found the friction on the trailer and plugged it into eq.1 and got a value for tension 3

Did the same to find out the friction of A plugged in value of T 2 did the same thing for truck.

Overall I have three values of tension now. The truck provides the tension of the value of T1.

Are my procedures correct?

There is 1 truck and 2 trailers in the problem. There are only two hitches. Why do you have 3 FTs? Be careful with your signs. If + is the direction the truck is moving, then some forces are negative. Each FT is either positive or negative, but not both. It is easiest to always make the unknowns positive when you set up the equations, and use the appropriate signs to indicate direction. For trailer A there are two tensions acting in oppostie directions, so they should have opposite signs, etc.