Calculating Velocity of 6000 lb Truck Around 650 ft Curve

  • Thread starter Thread starter cowgiljl
  • Start date Start date
  • Tags Tags
    Centripetal
Click For Summary

Homework Help Overview

The discussion revolves around calculating the velocity of a 6000 lb truck navigating a curve with a radius of 650 ft, considering the maximum friction force of 1125 lbs. Participants are exploring the relationships between force, mass, and velocity in the context of circular motion.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants present various calculations leading to different velocity results: 62.6 ft/sec, 11.04 ft/sec, and 58.88 ft/sec. There are discussions about the necessity of converting units from pounds to Newtons and the implications of using the metric system versus the imperial system. Some participants question the dimensional consistency of the calculations and the relevance of the maximum frictional force in determining the correct velocity.

Discussion Status

The conversation is ongoing, with some participants asserting that one of the answers is correct while others challenge the methods used to arrive at different results. There is no clear consensus, but several lines of reasoning are being explored regarding unit conversions and the application of physics principles.

Contextual Notes

Participants are working under the constraints of using the imperial system for this problem, which has led to confusion regarding unit conversions and the application of formulas. The urgency of the original poster's deadline for submission adds pressure to the discussion.

cowgiljl
Messages
63
Reaction score
1
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
 
Physics news on Phys.org
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.

[tex]F = \frac {mv^2}{r}[/tex]

[tex]\frac {Fr}{m} = v^2[/tex]

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

[tex]m = \frac {F}{a}[/tex]

[tex]m = \frac {6000}{32.2}[/tex]

[tex]m = 186.3[/tex] i think that's in slugs

now back to the other formula

[tex]v = \sqrt {\frac {Fr}{m}}[/tex]

[tex]v = \sqrt {\frac {(1125)(650)}{186.3}}[/tex]

[tex]v = 62.65[/tex]
 

Similar threads

  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
8K
Replies
7
Views
2K
Replies
1
Views
2K
Replies
40
Views
13K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 16 ·
Replies
16
Views
18K