A block of mass 4√2 kg is at rest on the inclined plane. The inclined plane is inclined at an angle of 135° from horizontal direction in anticlockwise direction. Determine the coefficient of friction so that block can start slide in downward direction, assume the acceleration of gravity as 10 m/s2.
Coefficient of friction must be less than 1
Coefficient of Friction on Inclined Plane
This problem asks for the condition on the coefficient of friction required for a block to start sliding downwards on an inclined plane. The block has a mass of \(4\sqrt{2}\) kg, and the plane is inclined at 135° from the horizontal in the anticlockwise direction.
Understanding the Angle of Inclination
The angle of inclination of the plane is given as 135° from the horizontal, measured anticlockwise. In typical inclined plane problems, the angle \(\theta\) used for force calculations is the angle the plane makes with the horizontal axis in the downward direction. If the angle is 135° anticlockwise from the positive x-axis (horizontal), the angle the plane makes with the horizontal measured towards the downward slope is \(180^\circ - 135^\circ = 45^\circ\). So, the effective angle of inclination for our calculations is \(\theta = 45^\circ\).
Forces Acting on the Block
When the block is on the inclined plane, the following forces act on it:
- Gravity (mg): Acts vertically downwards. Its magnitude is \(mg\).
- Normal Force (N): Acts perpendicular to the surface of the inclined plane, upwards.
- Friction Force (f): Acts parallel to the surface of the inclined plane, opposing the impending motion. Since the block is about to slide downwards, the static friction force acts upwards along the plane.
Resolving Gravity
We resolve the gravitational force into components parallel and perpendicular to the inclined plane. The angle between the vertical direction (gravity) and the normal to the plane (perpendicular to the plane) is equal to the angle of inclination, \(\theta = 45^\circ\).
- Component of gravity perpendicular to the plane: \(mg \cos\theta\)
- Component of gravity parallel to the plane (downwards): \(mg \sin\theta\)
Given mass \(m = 4\sqrt{2}\) kg, gravity \(g = 10\) m/s², and angle \(\theta = 45^\circ\).
Gravitational force parallel to the plane (downwards):
\[ F_{parallel} = mg \sin\theta = (4\sqrt{2})(10) \sin(45^\circ) \] \[ F_{parallel} = (40\sqrt{2}) \left(\frac{1}{\sqrt{2}}\right) \] \[ F_{parallel} = 40 \text{ N} \]Gravitational force perpendicular to the plane:
\[ F_{perpendicular} = mg \cos\theta = (4\sqrt{2})(10) \cos(45^\circ) \] \[ F_{perpendicular} = (40\sqrt{2}) \left(\frac{1}{\sqrt{2}}\right) \] \[ F_{perpendicular} = 40 \text{ N} \]Calculating Normal Force
The normal force \(N\) is perpendicular to the plane and balances the component of gravity perpendicular to the plane.
\[ N = F_{perpendicular} = 40 \text{ N} \]Condition for Starting to Slide Downward
The block will start to slide downward when the component of gravity parallel to the plane (downwards) becomes greater than the maximum static friction force acting upwards along the plane. The maximum static friction force is given by \(f_{s,max} = \mu_s N\), where \(\mu_s\) is the coefficient of static friction.
The condition for sliding downwards is:
\[ F_{parallel} > f_{s,max} \] \[ mg \sin\theta > \mu_s N \] \[ 40 > \mu_s (40) \]To find the condition on \(\mu_s\), we can divide both sides by 40:
\[ \frac{40}{40} > \mu_s \] \[ 1 > \mu_s \]So, the coefficient of static friction (\(\mu_s\)) must be less than 1 for the block to start sliding downwards.
Conclusion on Coefficient of Friction
For the block to begin moving downwards, the force pulling it down the incline must exceed the maximum possible static friction force opposing this motion. Our calculation shows this requires the coefficient of friction to be less than 1.
Let's compare this with the given options:
- Option 1: All the options are correct (Incorrect, as the condition is specific).
- Option 2: Coefficient of friction must be less than 1 (Matches our derived condition \(\mu_s < 1\)).
- Option 3: Coefficient of friction must be greater than 1 (Incorrect).
- Option 4: Coefficient of friction must be equal to 1 (Incorrect, equal to 1 means it is just on the verge of sliding, but needs to be overcome to *start* sliding).
Therefore, the coefficient of friction must be less than 1 for the block to start sliding downward.
Important Questions from Equilibrium and Friction
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Which of the following is not a type of equilibrium?
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A block of mass 10 kg is sliding on the ground with applied external force of 20 N. The coefficient of friction between the block and the ground is 0.1. Determine the linear acceleration of the block. Assume the acceleration due to gravity as 10 m/s2 .
