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Normal and Tangential Forces

As a particle moves around a curved path the equation of motion F=ma will need to modified to represent the tangential, normal, and binormal directions.  This can be represented by the equations below.

(Eq 1) $∑F_t = ma_t$

m = mass

at = tangential acceleration

ft = tangential force

(Eq 2) $∑F_n = ma_n$

an = normal acceleration

fn = normal force

(Eq 3) $∑F_b = 0$

$F_b$ = binormal Force

As a particle is moving through a curved path there is no motion of the particle in that direction.  Due to this fact there will be no resultant forces in the binormal direction.  There is however a possibility of a tangential force, and there will always be a normal force.  The tangential force is dependent on whether the tangential velocity is changing or if it is constant.  The normal force is there regardless of whether the tangential velocity is constant or not.  This due to the fact that a force is required to change the particles direction otherwise it would move in a straight line.  Recall that normal acceleration can be found by using the equation below.

(Eq 4)  $a_n=\frac{v_t^2}{r}$

The resultant force caused by tangential and normal forces can be found by using the equation below.

(Eq 5)  $F_R=sqrt{F_t^2+F_n^2}$

 

Example

As roller coaster goes around a corner normal acceleration will result.  Determine the bank of the corner to so that the trains normal force will be directed directly into the track so that the safety wheels aren’t be used to keep the train on the track.

Solution 

Step 1: Draw a free body diagram.

Step 2:  Using the n, b axes define the equations of motion.

(1) $∑F_n=ma_n$:  $F_csinθ=m\frac{v^2}{r}$

(2) $∑F_b=0$:  $F_ccosθ-mg=0$  or $F_ccosθ=mg$

Step 3: Divide (1)  by (2) and solve for θ

$\frac{F_csinθ=m\frac{v^2}{r}}{F_ccosθ=mg}=tanθ=\frac{v^2}{gr}$

$θ=tan^{-1}\left(\frac{v^2}{gv}\right)$

 

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