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Question: A truck of mass \(1000\,Kg\) is pulling a trailer of mass \(2000\,Kg\) frictional force on truck is ...

A truck of mass 1000Kg1000\,Kg is pulling a trailer of mass 2000Kg2000\,Kg frictional force on truck is 500N500\,N and on trailer is 1000N1000\,N. The truck extra force of 6000N6000\,N. Calculate:-
(A) Acceleration of the truck and the trailer.
(B) The tension in the connecting rope

Explanation

Solution

Use the formula of newton’s second law of motion and substitute the parameters of both truck and trailer to calculate the acceleration of the truck and the trailer. Then in order to calculate the tension in the connecting rope, total force acting on the trailer is considered.

Useful formula:
By the newton’s second law of motion
F=maF = ma
Where FF is the total force on the truck, mm is the mass of the truck and trailer and aa is the acceleration of the truck and the trailer.

Complete step by step solution:
Given data from the question are
Mass of the truck, M=1000KgM = 1000\,Kg
Mass of the trailer, m=2000Kgm = 2000\,Kg
Frictional force on the truck, FT=500N{F_T} = 500\,N
Frictional force on the trailer,Ft=1000N{F_t} = 1000\,N
Total force on the truck, F=6000NF = 6000\,N


(A) By considering the formula of newton’s second law,
F=maF = ma
mass is taken as the sum of the mass of truck and trailer
F=(M+m)aF = \left( {M + m} \right)a
Substituting the values of masses and force in the above equation.
6000(500+1000)=(1000+2000)a6000 - \left( {500 + 1000} \right) = \left( {1000 + 2000} \right)a
In the above equation, frictional force is subtracted from the total force. This is because some force is lost through friction.
60001500=3000a6000 - 1500 = 3000a
4500=3000a4500 = 3000a
By further simplification,
a=1.5ms1a = 1.5\,m{s^{ - 1}}.
Thus the acceleration of truck and trailer is obtained as 1.5ms11.5\,m{s^{ - 1}}.

(B) Let us consider that T be the tension of the connecting rope that connects the truck and the trailer.
Tension in the rope is the sum of the force acting on the trailer and the frictional force acting on it.
T=Ft+forceontrailerT = {F_t} + force\,on\,trailer
T=1000+2000aT = 1000 + 2000a
T=(2000×1.5)+1000T = \left( {2000 \times 1.5} \right) + 1000
By doing arithmetic operation in the above equation,
T=4000NT = 4000\,N

Tension in the connecting rope is obtained as 4000N4000\,N

Note: For calculating the tension acting on the rope that connects the truck and the trailer, only the total force that acts on the trailer is considered. This is because tension is the horizontal force , and this force in the rope is mainly given by the trailer as it is pulled by it.