Calculating Deceleration and Force: Car and Trailer Braking Forces

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SUMMARY

The discussion focuses on calculating the deceleration and horizontal force exerted by a trailer on a car during braking. A car weighing 1200 kg pulls a trailer weighing 1050 kg, both moving at 90 km/h. The braking forces are 4500 N for the car and 3600 N for the trailer. The calculated deceleration of the combined system is 3.6 m/s², and the horizontal force exerted by the trailer on the car is determined to be 180 N.

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Apprentice123
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A car of 1200 kg pull a trailer of 1050 kg. The set is moving at 90 km/h suddenly suffers the action of the brakes. Knowing that the forces of officers in the car brakes on the trailer and worth 4500N and 3600N, respectively, determine (a) the deceleration of the joint and (b) the horizontal component of force exerted by the trailer on the car

I think:

Car

ZFx = mc . a
-Ftc = mc . a

Trailer

ZFx = mt . a
Ftc = mt . a


But not possible to calculate
 
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Apprentice123 said:
A car of 1200 kg pull a trailer of 1050 kg. The set is moving at 90 km/h suddenly suffers the action of the brakes. Knowing that the forces of officers in the car brakes on the trailer and worth 4500N and 3600N, respectively, determine (a) the deceleration of the joint and (b) the horizontal component of force exerted by the trailer on the car

I think:
Car
ZFx = mc . a
-Ftc = mc . a

Trailer
ZFx = mt . a
Ftc = mt . a

But not possible to calculate

What is the deceleration of the combined system?

∑F = (∑m)*a

a = ∑F/∑m

Then, knowing the deceleration of the whole, draw a force diagram to figure the forces between the parts.
 
LowlyPion said:
What is the deceleration of the combined system?

∑F = (∑m)*a

a = ∑F/∑m

Then, knowing the deceleration of the whole, draw a force diagram to figure the forces between the parts.

Thank you.

a = (4500+3600)/(1200+1050) = 3,6 m/s^2

Break + F = mB . a
F = (1050 x 3,6) - 3600
F = 180 N
 

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