Problem - Newton's second law applied to uniform circular motion

opticaltempest

Here is the problem I am working on:



Here is my work:



I cannot solve (2) for v since there are no real roots.

When I set up the net forces in the x-direction in (1), should I have made [tex]ma_c[/tex] negative? I think it makes sense if [tex]ma_c[/tex] is negative because it tells us that the acceleration vector is always pointing in the negative x-direction.

In all of the other Newton's second law problems, I never had to make the side with [tex]ma[/tex] negative. Why in this problem must I now specify the sign of the acceleration?

Doc Al

The minus sign just signifies direction. If the friction is negative, so is the acceleration. [itex]v^2/r [/itex] gives the magnitude of the acceleration, not the direction.

radou

Unless I'm missing something, the friction should point in the opposite direction, which is the answer to your problem. The force of friction always has a direction opposite to the direction of motion (or, in this case, to the direction of a force which would cause such motion).

Chi Meson

In circular motion, the acceleration does NOT always point in any direction. The centripetal accelration always points toward the center of the circle, hence the direction keeps changing. Just toss the negative anyhow, since you are only solving for the magnitude. When informing the direction, all that's needed is "toward the center of the circle."

opticaltempest

Radou:

In this problem I assumed the truck was driving in a counterclockwise circle, therefore the force preventing the truck from moving in a straight line was the inward pointing [tex]f_s[/tex].

Doc Al and Chi Meson:

I think that cleared up my mistake.

So, on these specific types of problems I should not worry about including the direction of the static friction force and the direction of the centripetal acceleration in the plane containing the circle (since we already know the static friction force vector and centripetal acceleration vector always points inward at every point along the circle assuming we have uniform circular motion / constant speed)?

Doc Al

If two vectors point in the same direction, they are either both negative or both positive (depending upon your choice of coordinate system). Either way, the signs cancel. If left is negative, and the acceleration points left, then the centripetal acceleration (including proper sign) is [itex]-v^2/r[/itex], not [itex]v^2/r[/itex].