Centripetal accel

1. Dec 16, 2003

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 excert 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

2. Dec 16, 2003

himanshu121

B) is correct
i dont know how u reached the other results Pls show it

3. Dec 17, 2003

harsh

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.

4. Dec 17, 2003

himanshu121

B is correct harsh u are given max frictional force why u are dividing by 9.81

5. Dec 17, 2003

HallsofIvy

Staff Emeritus
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?

6. Dec 17, 2003

harsh

Right, but then you dont really get the correct units.
Its like you are canceling pounds over pounds, so the left side of the equation is unitless.

7. Dec 17, 2003

himanshu121

It should be 1125lbs m/s^2

8. Dec 19, 2003

ShawnD

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

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

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

$$v = 62.65$$