Question

If the horizontal effort \(P\) is applied to a body of weight \(W\) and is about to move up on the rough inclined plane whose angle of inclination with the horizontal is \(\alpha\), then the effort \(P=\) (Where \(\mu=\) coefficient of friction between the plane and the body, \(\phi=\) friction angle)

• A. \(P=W\tan\alpha\)
• B. \(P=W\tan(\alpha+\phi)\)
• C. \(P=W(\sin\alpha+\mu\cos\alpha)\)
• D. \(P=W(\cos\alpha+\mu\sin\alpha)\)

Hint:

For rough inclined plane problems: \[ \tan\phi=\mu. \] If the body is just moving up the plane, \[ \boxed{P=W\tan(\alpha+\phi)} \] whereas for impending motion down the plane, \[ \boxed{P=W\tan(\alpha-\phi).} \] These are standard results frequently used in Engineering Mechanics.

Answer 1

The Correct Option is B

Concept: For a body on a rough inclined plane that is just about to move upward under the action of a horizontal force, the limiting friction acts downward along the plane. Using the principle of limiting equilibrium and the angle of friction, \[ \tan\phi=\mu, \] the required horizontal effort is \[ \boxed{P=W\tan(\alpha+\phi).} \] This is a standard result in Engineering Mechanics.

Step 1: Identify the direction of friction. Since the body is about to move upward, - Friction acts downward along the plane. - The body is in limiting equilibrium. The angle of friction satisfies \[ \tan\phi=\mu. \]

Step 2: Apply the equilibrium condition. Resolving the forces along and perpendicular to the inclined plane and using the limiting friction condition, \[ F=\mu N, \] the horizontal effort required is obtained as \[ P=W\tan(\alpha+\phi). \] Hence, \[ \boxed{P=W\tan(\alpha+\phi).} \] Therefore, the correct option is \[ \boxed{(B)\;W\tan(\alpha+\phi).} \]