Will the Tow Cable Hold During the Car Tow?

Click For Summary

Homework Help Overview

The discussion revolves around a scenario where a tow truck is attempting to tow a broken-down car. The problem involves analyzing the forces acting on the car, including the tension in the tow cable, the weight of the car, and the effects of kinetic friction, given specific parameters such as the weights of the vehicles and the angle of the cable.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to calculate the sum of forces in the x direction and how to incorporate kinetic friction into their analysis. There are considerations of using Newton's second law and the effects of the angle of the cable on the forces involved.

Discussion Status

Some participants have provided insights into the equations governing the forces, while others express uncertainty about the initial steps and the incorporation of friction. There is an ongoing exploration of the relationships between tension, friction, and acceleration without a clear consensus on the final outcome.

Contextual Notes

Participants are working under the assumption that the rolling friction behaves like kinetic friction, with a specified coefficient. The discussion also reflects on the limitations of the cable's strength and the implications of the angle at which it is attached.

bignasty
Messages
10
Reaction score
0
The Problem:
Your car breaks down in the middle of nowhere. A tow truck weighing 4000 lbs. comes along and agrees to tow your car, which weighs 2000 lbs., to the nearest town. The driver of the truck attaches his cable to your car at an angle of 20 degrees to horizontal. He tells you that his cable has a strength of 500 lbs. He plans to take 10 secs to tow your car at a constant acceleration from rest in a straight line along a flat road until he reaches the maximum speed of 45 m.p.h. Can the driver carry out the plan? You assume that the rolling friction behaves like kinetic friction, and the coefficient of the rolling friction between your tires and the road is 0.10.

I'm not sure where to start. I'm pretty sure I need to find the sum of all the forces in the x direction, then use the angle of 20 degrees to see if the the tension on the rope is more that 500 lbs. Mainly, I am just not sure how to incorporate the kinetic friction. If anyone could provide a hint as to how incorporate the kinetic friction, or just let me know if I'm even on the right track, I'd appreciate it. Thanks
 
Physics news on Phys.org
The force from kinetic friction is anti parrallel to the direction of the car's motion.

The equation for the magnitude of this force

F_f = \mu_k n

where n is the normal force and u_k is the coefficient of friction.

your definitely on the right track, consider Newton's second equation for horizontal and vertical components of force on the car.
 
bignasty said:
The Problem:
Your car breaks down in the middle of nowhere. A tow truck weighing 4000 lbs. comes along and agrees to tow your car, which weighs 2000 lbs., to the nearest town. The driver of the truck attaches his cable to your car at an angle of 20 degrees to horizontal. He tells you that his cable has a strength of 500 lbs. He plans to take 10 secs to tow your car at a constant acceleration from rest in a straight line along a flat road until he reaches the maximum speed of 45 m.p.h. Can the driver carry out the plan? You assume that the rolling friction behaves like kinetic friction, and the coefficient of the rolling friction between your tires and the road is 0.10.

I'm not sure where to start. I'm pretty sure I need to find the sum of all the forces in the x direction, then use the angle of 20 degrees to see if the the tension on the rope is more that 500 lbs. Mainly, I am just not sure how to incorporate the kinetic friction.
I assume that the tension limit is 500 lb force (ie the weight of 500 lb = 500*32 = 16000 ftlb of force).

The forces on the tow cable are friction, gravity and ma. (note: the vertical component of the tension reduces the normal force which reduces friction).

F_{acceleration} = ma
F_{friction} = \mu_kN = \mu_k(mg - Tsin(20))

So:
(1)Tcos(20)=F_{acceleration} + F_{friction} = ma + \mu_k(mg-Tsin(20))

Tcos(20) = ma + \mu_kmg - \mu_kTsin(20))
T(cos(20)+sin(20)\mu_k) = ma + \mu_kmg

(2)T = \frac{ma + \mu_kmg}{(cos(20)+sin(20)\mu_k)}

ma = m\Delta v / \Delta t = 2000 * 66/10 = 13200lbft/sec^2 (45 mph = 66 ft/sec)

T = \frac{13200 + .1*2000*32}{(.94 + .34*.1)}

T = \frac{13200 + 6400}{.974} = 19770 > 16000lb ft/sec^2

AM
 
Last edited:
Thanks a lot for the explanations!

bignasty
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
7K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
12
Views
2K
  • · Replies 9 ·
Replies
9
Views
5K
Replies
2
Views
8K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
8K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 10 ·
Replies
10
Views
5K