Calculating Velocity of 6000 lb Truck Around 650 ft Curve

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SUMMARY

The discussion focuses on calculating the velocity of a 6000 lb truck navigating a 650 ft curve, with a maximum friction force of 1125 lbs. The participants derived three different velocity results: 62.6 ft/sec, 11.04 ft/sec, and 58.88 ft/sec. The correct calculation involves using the formula F = mv²/r, converting the friction force to Newtons, and determining the mass in slugs, leading to a final velocity of approximately 62.65 ft/sec. The consensus is that the first answer is accurate based on proper unit conversions and application of physics principles.

PREREQUISITES
  • Understanding of Newton's laws of motion (F = ma)
  • Knowledge of circular motion dynamics (Ff = mv²/r)
  • Familiarity with unit conversions between pounds and Newtons
  • Basic grasp of mass calculations in the imperial system (slugs)
NEXT STEPS
  • Study the principles of circular motion and centripetal force calculations
  • Learn about unit conversion techniques between imperial and metric systems
  • Explore the implications of friction in vehicle dynamics
  • Review examples of real-world applications of F = ma in engineering contexts
USEFUL FOR

This discussion is beneficial for physics students, automotive engineers, and anyone interested in understanding vehicle dynamics and frictional forces in motion.

cowgiljl
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Truck weighs 6000 lbs The truck is going around a curve with a radius of 650 ft. The max friction force the road can exert is 1125 lbs.Its velocity needs to be calculated ft/sec.

Now I now F=ma and Ff = m*(v^2/r)
I also now that 1 pound = 4.448 N

problem is i have 3 different answers by trying it different ways

a) v = 62.6 ft/sec
b) v = 11.04 ft/sec
c) v = 58.88 ft/sec

Which one is the right one

This is the final problem I have to do before i have to turn in my physics notebook and take my final exam today at 9 am

any assistance would be great

thanks
 
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B) is correct
i don't know how u reached the other results Pls show it
 
I think there is a mistake in dimension somewhere in there. Actually, the frictional force must be converted to Newtons and the weight of the truck must be converted to kg. Then, you can use [

f = ma
f = v^2 / r

(1125 lbs)*(4.448 N/lbs) = [(6000 lbs * 4.448 N/lbs)/9.81 m/s^2]*v^2/r

You will get around 34 ft/sec.
 
B is correct harsh u are given max frictional force why u are dividing by 9.81
 
Actually, the frictional force must be converted to Newtons and the weight of the truck must be converted to kg.

Why? You are given all the information in the "English" system and asked for the result in the "English" system. F= ma and F= mv2/r are true in any system. Why convert to metric?
 
Right, but then you don't really get the correct units.
Its like you are canceling pounds over pounds, so the left side of the equation is unitless.
 
The max friction force the road can exert is 1125 lbs

It should be 1125lbs m/s^2
 
Originally posted by cowgiljl
Truck weighs 6000 lbs The truck is going around a curve with a radius of 650 ft. The max friction force the road can exert is 1125 lbs.Its velocity needs to be calculated ft/sec.

Now I now F=ma and Ff = m*(v^2/r)
I also now that 1 pound = 4.448 N

problem is i have 3 different answers by trying it different ways

a) v = 62.6 ft/sec
b) v = 11.04 ft/sec
c) v = 58.88 ft/sec

Which one is the right one

I'm getting the first answer. I've never actually used the imperial system before; it's just weird.

F = \frac {mv^2}{r}

\frac {Fr}{m} = v^2

we have to find the mass of the thing. divide the weight by the acceleration to get the mass.

m = \frac {F}{a}

m = \frac {6000}{32.2}

m = 186.3 i think that's in slugs

now back to the other formula

v = \sqrt {\frac {Fr}{m}}

v = \sqrt {\frac {(1125)(650)}{186.3}}

v = 62.65
 

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