Question

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Answer 1

Answer:

umgcos© + mgsin© = F

put mg = 1

Answer 2

Solution:

Weight of object along the inclined plane = W sinФ

W = 1 N  (Given)

when blocks moves upward, friction will be in the downward direction.

Fr = μR

R = normal reaction = W cos Ф

Fr = μ sin Ф

Net upward force = net downward force

So, applied force in downward direction = sin Ф + μ cos Ф

Hence, 4). option is correct.

Extra Info:

Motion up an inclined plane:

Note: Diagram attached in attachment file.

On y axis:

R + F sin β = mg cos Ф       .............(1)

On x axis:

mg sin Ф + Fs = F cos β     .............(2)

Fs = μR = μ(mg cos Ф - F sin β)        ...............(3)

mg sin Ф + μ mg cos Ф - μF sin β = F cos β

mg (sin Ф + μ cos Ф) = F cos β + μ sin β)

$\therefore {\boxed{\boxed{\sf{F = \dfrac{mg \sin \theta + \mu mg \cos \theta}{\cos \beta+ \mu \sin\beta}}}}}$

μs = tan Фs

$\sf{F = \dfrac{mg \sin \theta + mg \tan \theta s\cos \theta}{\cos \beta+ \tan \theta s. \sin\beta}}$

$\sf{=\dfrac{mg\bigg(\sin \theta + \dfrac{\sin \theta s}{\cos \theta s}.\cos \theta\bigg)}{\cos \beta+\dfrac{\sin \theta s}{\cos \theta s}.\sin\beta}}$

$\sf{\implies F = \dfrac{mg (\cos\theta s. \sin\theta + \sin\theta s. \cos\theta)}{\cos \beta .\cos \theta s+ \sin\theta s .\sin\beta}}$

${\boxed{\boxed{\sf{\implies F = \dfrac{mg.\sin(\theta+\theta s)}{\cos(\beta-\theta s}}}}}$

From F to be min, cos (β - Фs) = 1

∴ β - Фs = 0

∴ β = Фs

Fmin = mg . sin (Ф + Фs)