Animated Solution for Physics - Laws of Motion: The normal reaction $N$ for a vehicle of $800\text{ kg}$ mass, negotiating a turn on a $30^{\circ }$ banked road at maximum possible speed without skidding is $\cdots \times 10^3{\text{ kg-m/s}}^2$. [Take, ${\mu }_s=0.2$]

A Shortcut to Normal Reaction

The banked curve is a classic physics problem that often leads students into a tangled web of sine and cosine equations. When a vehicle takes a turn on a banked road, it experiences a complex interplay of gravity, normal reaction, and friction.

Most textbooks teach you to resolve forces parallel and perpendicular to the incline, or horizontally and vertically, and then solve a system of simultaneous equations. But what if I told you there's a much simpler, more elegant way to find the normal reaction? Let's dive in.

Imagine a vehicle of mass $m=800\text{ kg}$ negotiating a turn on a road banked at an angle $\theta =30^{\circ }$. The question asks for the normal reaction $N$ when the vehicle is traveling at the maximum possible speed without skidding.

First, we must identify the forces acting on the vehicle: 1. Weight ($mg$): Acts vertically downwards. 2. Normal Reaction ($N$): Acts perpendicular to the road surface, pushing the car up and inwards. 3. Static Friction ($f_s$): This is the crucial part. At maximum speed, the car has a tendency to skid outwards (up the incline). To oppose this impending motion, static friction must act down the incline.

Here is the brilliant shortcut. Instead of worrying about the horizontal centripetal acceleration ($a_c=\frac{v^2}{R}$), let's look exclusively at the vertical direction.

Is the car accelerating vertically? Is it flying into the air or sinking into the asphalt? No. The vehicle remains on the horizontal plane of its circular path. This means the net vertical force must be exactly zero.

Let's balance the vertical forces: Upward Force: The vertical component of the normal reaction, which is $N\cos 30^{\circ }$. Downward Forces: The weight of the car $mg$, and the vertical component of the friction force, which is $f_s\sin 30^{\circ }$.

Equating the upward and downward forces gives us our master equation:

We know that at the maximum speed, the static friction is at its limiting value. Therefore, $f_s={\mu }_{s}N$. Let's substitute this into our equation:

Now, it's just simple algebra to isolate $N$. Bring all terms containing $N$ to the left side:

Factor out $N$:

Finally, solve for $N$:

Look at how beautiful that is! We found an expression for the normal reaction without ever needing to know the velocity $v$ or the radius of the turn $R$.

Now, let's plug in the given values: $m=800\text{ kg}$, $\theta =30^{\circ }$, and ${\mu }_s=0.2$.

A quick tip for JEE: When the value of $g$ is not explicitly given, look at the options. If we use $g=10{\text{ m/s}}^2$, the calculation yields approximately $10.44\times 10^3\text{ N}$, which doesn't perfectly match any option. However, using $g=9.8{\text{ m/s}}^2$ will lead us right to the target.

Expressing this in the format requested by the question:

This perfectly matches option (a). By understanding the physical constraints (zero vertical acceleration) and the direction of impending motion, we bypassed the messy simultaneous equations and arrived at the solution with elegance and speed.