Find tension if given friction and acceleration?

Question

A 2660kg car is being towed in a straight line by a tow truck. The car has its emergency brake and on. The coeifficent of friction is .540 and both the car and truck are traveling at 2.50 m/s^2. What is the tension in the cable between the vehicles.
Assume the cable is weightless.

Answer 1

Wellllllllllll this was a good question where i think you could hav solved it by a little intelligence.

So here goes

1) Draw FBD
2)Write the eqautions. which are

For Truck
Force - Friction- tension=mass of truck * accn

for car
Tension - friction = mass of car * accn
T= (2660*2.5) + ( .54*2660*9.8)

I hope you have a calculator at home so no calculation from my side.

Answer 2

i believe we need the mass of the tow truck to solve this problem.
1st assumption: the car hits the brake and it's deceleration is 2.50 m/s^2
say the truck has a mass of 5000kg
the friction force on the car is 2660 x .54 x 9.81 = 14091 N
the acceleration of the truck remains at 2.50 m/s^2 at the moment. so the force on the truck is 5000 x 2.50 = 12500 N
so the force in the cable HAS TO BE 12500 N.
if the truck has a mass of 10000kg, then the force on the truck will be 25000 N and the force in the cable will be 14091 N (the car will skid).
think, if you can't figure this out immediately. cheers! =)

Answer 3

Tension in the cable is equal to the force of friction acting on the tires of the car. f=uN, where f equals the force of friction, u the coefficient of friction, and N the force acting downwards on the tires expressed in Newtons.

1 kg of mass =9.8 Newtons
N=2660*9.8 Newtons.

f=0.540*2660*9.8
= 14,077 Newtons

Answer 4

2660*.540*9.8N
the drag produced by the car is independant of speed so that the tension on the cable
is constant.

Answer 5

Ff = μmg
af = μg
F = m(a+μg)
F = 2,660(2.5 + 0.54*9.8)
F = 20,726.72 N