Question

Suppose that you are standing on a train accelerating at 0.20g. What minimum coefficient of static friction must exist between your feet and the floor if you are not to slide.

Answer 1

Acceleration = (0.2 x g) = 1.96m/sec^2. 
Accelerating force on 1kg. = (ma) = 1.96N. 
1kg. has a weight (normal force) of 9.8N. 
Coefficient µ = 1.96/9.8 = 0.2 minimum. 

Coefficient is a ratio, so holds true for any value of mass to find accelerating force acting. 
e.g. 75kg = (75 x g) = 735N. 
Accelerating force = (735 x 0.2) = 147N

Answer 2

The coefficient of static friction between the feet and the floor is $\boxed{0.2}$.

Further explanation:

Here, we have to calculate the coefficient of static friction between the feet and the floor.

Given:

The acceleration of the train is $0.2g$.

Concept:

As we know, the friction occurs between the two surface when there is a relative motion of the one surface or body over the other surface.

Maximum friction force is the limiting friction and its value is equal to the product of the coefficient of the static friction and the normal reaction.

So, friction force can be calculated as,

$\boxed{{F_r}=\mu mg}$

Here, $m$ is the mass of the observer, $\mu$ is the coefficient of static friction,

Here, according to thequestion, in the given condition, the observer (you) will not slide until observer try to slide.

Therefore, if the observer does not slide relative to the surface of the train, it means that the acceleration of the observer is same as that of the acceleration of the train.

So, acceleration of the observer will be equal to the acceleration of the train.

$\boxed{a=0.2g}$

Here, $a$ is the acceleration of the observer (you).

We know that, when an object accelerates then the inertia force experienced by the object in the opposite direction of the acceleration.

So, inertia force will be,

${F_I} = ma$

So, by applying the equilibrium condition,

${F_r} = {F_I}$

Substitute the value of ${F_r}$ and ${F_I}$ in the above equation.

$\begin{aligned}\mu mg&=ma\\\mu mg&=m\left( {0.2g} \right)\\\mu g&=0.2g\\\mu&=0.2\\\end{aligned}$

Thus, the coefficient of static friction between the feet and the floor is $\boxed{0.2}$.

Learn more:

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Answer detail:

Grade: Senior School

Subject: Physics

Chapter: Kinematics

Keywords:

Standing, train, accelerating, coefficient of the static friction, not slide, mg, umg, moving, relative motion, surface, friction force.