Calculating Deceleration and Force: Car and Trailer Braking Forces

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In summary, a car of 1200 kg pulling a trailer of 1050 kg is traveling at 90 km/h when the brakes are suddenly applied. The forces of officers in the car brakes and trailer are 4500N and 3600N, respectively. It is not possible to calculate the deceleration of the joint or the horizontal component of force exerted by the trailer on the car. However, using the formula ∑F = (∑m)*a, the deceleration of the combined system can be found to be 3.6 m/s^2. Drawing a force diagram and using the formula Break + F = mB . a, the horizontal component of force exerted by the trailer on the
  • #1
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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  • #2
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.
 
  • #3
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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