Calculating Force Exerted by 1000kg Car on 450kg Trailer

  • Thread starter Thread starter dabouncerx24
  • Start date Start date
  • Tags Tags
    Car Force Trailer
AI Thread Summary
A 1000kg car exerts a horizontal force of 3500N to pull a 450kg trailer, and the frictional force for the trailer, calculated using a coefficient of 0.15, must be considered in determining the force exerted on the trailer. The frictional force is derived from the equation (0.15)(450)(9.81), which needs to be either added to or subtracted from the car's force. The discussion also shifts to a separate problem involving a car's skid mark of 88m on a rainy day with a friction coefficient of 0.42, prompting a calculation of the car's speed at the moment brakes were applied. It is noted that the car's mass is irrelevant in this context due to the cancellation of mass in the equations used. Understanding these principles is crucial for solving both problems effectively.
dabouncerx24
Messages
11
Reaction score
0
The problem is, a 1000kg car pulls a 450kg trailer. The car exerts a horizontal force of 3500N against the ground in order to accelerate. What force does the car exert on the trailer? Assume a friction coefficient of 0.15 for the trailer.
 
Physics news on Phys.org
dabouncerx24 said:
The problem is, a 1000kg car pulls a 450kg trailer. The car exerts a horizontal force of 3500N against the ground in order to accelerate. What force does the car exert on the trailer? Assume a friction coefficient of 0.15 for the trailer.

You need to either add the frictional force to the 3500N or subtract it. The frictional force equals (0.15)(mg) or (0.15)(450)(9.81).
 
I don't get why adding the friction force to 3500N is logical? Can you please explain it to me.
 
Do the forces analysis...
 
Well I figured out that problem.

Here is another one, please assist me.

At an accident scene on a level road, investigators measure a car;s skid mark to be 88m long. It was a rainy day and the coefficient of friction was estimated to be 0.42. Use these dta to determine the speed of the car when the driver slammed on the brakes (why does the car's mass not matter).

So far I figured I would use Fd=KE-KE' so all the ms will cancel each other.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top