Question

A block of mass m is lying on an inclined plane. The coefficient of friction is \(\mu\). The force required to move the block up the inclined plane will be

• A. \(mg \sin \theta - \mu mg \cos \theta\)
• B. \(mg \sin \theta + \mu mg \cos \theta\)
• C. \(mg \cos \theta - \mu mg \sin \theta\)
• D. \(mg \cos \theta + \mu mg \sin \theta\)

Hint:

When an object is pushed up an incline, both the parallel component of gravity (\(mg \sin \theta\)) and friction (\(\mu mg \cos \theta\)) act in the same direction (down the incline) and must be overcome. Thus, they add up. If the object were sliding down, friction would act up the incline, opposing the gravitational component.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the minimum force required to push a block up an inclined plane, overcoming both gravity and friction.

Step 2: Key Formula or Approach:
We will use a free-body diagram and apply Newton's First Law (for the condition of impending motion, acceleration is zero). The main forces are the applied force, gravity, normal force, and friction.

Step 3: Detailed Explanation:
Let's analyze the forces acting on the block along axes parallel and perpendicular to the inclined plane.
1.

Gravitational Force (Weight): \(mg\), acting vertically downwards. - Component parallel to the incline: \(mg \sin \theta\) (acting down the incline). - Component perpendicular to the incline: \(mg \cos \theta\) (acting into the incline).
2.

Normal Force (N): Acts perpendicular to the surface, outwards. From equilibrium in the perpendicular direction, \(N = mg \cos \theta\).
3.

Frictional Force (f): Opposes the motion (or impending motion) up the plane, so it acts down the plane. The maximum static friction (or kinetic friction) is \(f = \mu N = \mu mg \cos \theta\).
4.

Applied Force (F): The force required to move the block up the plane, acting parallel to the incline, upwards.
For the block to move up, the applied force F must overcome the sum of the forces pulling it down the incline.
\[ F = (\text{Gravitational component down the incline}) + (\text{Frictional force down the incline}) \]
\[ F = mg \sin \theta + f \]
Substitute \(f = \mu mg \cos \theta\):
\[ F = mg \sin \theta + \mu mg \cos \theta \]

Step 4: Final Answer:
The force required to move the block up the inclined plane is \(mg \sin \theta + \mu mg \cos \theta\).