Question

A car moving at a speed of 36 km/hr is taking a turn on a circular road of radius 50m. A small plate is kept on the seat with its plane perpendicular to the radius of the circular road in figure. A block of mass 100g is kept on hte seat which rests against the plate. The friction coefficient be the block and the plate is µ = 0.58.

(a) Find the normal contact force exerted by the plate on the block. (b) The plate is slowly turned so that the angle between the normal to the plate and the radius road slowly increases. Find the angle at which the block will just start sliding on the plate.

Answer 1

(a) Normal contact force exerted by the plate on the block = 0.2 N (b) The angle at which the block will just start sliding on the plate = 30.1 degrees

Answer 2

The problem describes a car moving on a circular road, with a block resting against a plate on the seat. We need to find the normal force exerted by the plate on the block in the initial setup and then the angle at which the block starts sliding when the plate is turned.

Given values:

• Speed of car, $v = 36 \text{ km/hr}$
• Radius of circular road, $R = 50 \text{ m}$
• Mass of block, $m = 100 \text{ g}$
• Coefficient of friction, $\mu = 0.58$

Step 1: Convert units to SI units.

• Speed $v = 36 \text{ km/hr} = 36 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 10 \text{ m/s}$.
• Mass $m = 100 \text{ g} = 0.1 \text{ kg}$.

Part (a): Find the normal contact force exerted by the plate on the block in the initial setup.

In the initial setup, the plane of the plate is perpendicular to the radius of the circular road. This means the normal to the plate points directly towards the center of the circular path. For the block to move in a circle along with the car, a centripetal force is required, directed towards the center. This centripetal force is provided by the normal contact force ($N$) from the plate on the block.

The formula for centripetal force is $F_c = \frac{mv^2}{R}$. So, the normal force $N = F_c = \frac{mv^2}{R}$.

Substitute the values: $N = \frac{(0.1 \text{ kg}) \times (10 \text{ m/s})^2}{50 \text{ m}}$ $N = \frac{0.1 \times 100}{50}$ $N = \frac{10}{50}$ $N = 0.2 \text{ N}$

Part (b): Find the angle at which the block will just start sliding on the plate.

When the plate is slowly turned, the normal to the plate makes an angle $\theta$ with the radial direction (direction towards the center of the circular path).

Let's analyze the forces in the frame of reference of the car (a non-inertial frame). In this frame, the block experiences a centrifugal force ($F_{cf}$) directed radially outwards. $F_{cf} = \frac{mv^2}{R}$.

The block is also subject to gravity ($mg$) downwards and a normal force from the seat ($N_{seat}$) upwards. These vertical forces balance each other and do not affect the horizontal motion or sliding against the plate.

Consider the forces in the horizontal plane: The normal force ($N'$) from the plate acts perpendicular to the plate. The friction force ($f_s$) acts parallel to the plate's surface.

The centrifugal force $F_{cf}$ acts radially outwards. When the plate is turned by an angle $\theta$, the normal to the plate makes an angle $\theta$ with the radial direction. We resolve the centrifugal force $F_{cf}$ into two components relative to the plate:

1. Component perpendicular to the plate (along the normal $N'$): This component is balanced by the normal force $N'$ exerted by the plate. $N' = F_{cf} \cos \theta = \frac{mv^2}{R} \cos \theta$.
2. Component parallel to the plate: This component tends to make the block slide along the plate. This component is $F_{cf} \sin \theta = \frac{mv^2}{R} \sin \theta$. To prevent sliding, the static friction force ($f_s$) must oppose this component. So, $f_s = \frac{mv^2}{R} \sin \theta$.

The block will just start sliding when the static friction force reaches its maximum possible value, which is $f_{s,max} = \mu N'$.

Set $f_s = f_{s,max}$: $\frac{mv^2}{R} \sin \theta = \mu N'$

Substitute the expression for $N'$: $\frac{mv^2}{R} \sin \theta = \mu \left( \frac{mv^2}{R} \cos \theta \right)$

We can cancel $\frac{mv^2}{R}$ from both sides (since it's non-zero): $\sin \theta = \mu \cos \theta$

Divide by $\cos \theta$: $\tan \theta = \mu$

Now, substitute the given value of $\mu$: $\tan \theta = 0.58$

To find $\theta$, we take the inverse tangent: $\theta = \arctan(0.58)$

Using a calculator, $\arctan(0.58) \approx 30.11^\circ$. So, $\theta \approx 30.1^\circ$.

The angle at which the block will just start sliding on the plate is approximately $30.1^\circ$.