Question 1198338
.
How much force is required to keep a 3.000 lb car 
travelling at 45 mph from skidding on the curve of radius of 400 ft.
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<pre>
They want you find the magnitude of the centripetal force.


For it, use the formula

    {{{F[centripetal]}}} = {{{(m*v^2)/R}}},


where m is the mass, v is the linear speed, R is the radius, in consistent units.


Convert 45 mph to feet per second  45 mph = {{{(45*5280)/3600}}} = 66 ft/s.


Convert 3000 lbs to mass by dividing the weight by the gravity acceleration g = 32 ft/s^2

    m = {{{3000/32}}} lbs/(ft/s^2).


Now substitute the values into the formula

    {{{F[centripetal]}}} = {{{((3000/32)*66^2)/400}}} = 1021 lbs.    <U>ANSWER</U>


If you ask which force provides it, &nbsp;I will answer you : &nbsp;partly the friction force 
and partly the gravity &nbsp;(if the road has a necessary slope to the center of the curve).
</pre>

Solved.


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<H3>The last post-solution notice</H3>

 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;these units, &nbsp;lbs, &nbsp;feet and miles, &nbsp;are hopelessly outdated for use in tasks in &nbsp;Physics.


 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It is just anti-pedagogic and anti-humane to use them in nova days.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;I would say, it is just already beyond the red line 
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;that separates educated people from their antipodes.


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An addition, added after viewing the post by Alan.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;After the year of &nbsp;1960, &nbsp;when &nbsp;SI &nbsp;system 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;was officially established and units &nbsp;" meter ", &nbsp;" kilogram " &nbsp;and  "&nbsp;newton " 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;were introduced for the length, &nbsp;mass and force, &nbsp;respectively,

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;no one textbook/(problems book) &nbsp;in &nbsp;Physics uses the old units &nbsp;" foot ", &nbsp;" mile " &nbsp;and &nbsp;" pound ".