Question

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Resolving must explained neatly (Diagram).

In the given figure , wedge is acted upon a constant horizontal force F .The wedge is moving on smooth horizontal surface. A ball of mass "m" is rest relative to wedge . The ratio of forces exerted on "m" by wedge when' F' is acting and 'F' is withdrawn assuming on friction between the edge and ball is equal to
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Answer 1

Explanation:

Question :

• In the given figure , wedge is acted upon a constant horizontal force F .The wedge is moving on smooth horizontal surface. A ball of mass "m" is rest relative to wedge . The ratio of forces exerted on "m" by wedge when' F' is acting and 'F' is withdrawn assuming on friction between the edge and ball is equal to

given :

• constant horizontal force = F

• A ball of mass = m

• ratio of forces exerted on "m" by wedge = F

to find :

• assuming on friction between the edge and ball is equal to

solution :

• please check the attached file. where you find full answer

• please check the attached file
• The ball moves along x- -axis with an acceleration, say a, whereas it is equilibrium along y axis

• When F = Next, the ball moves down the plane, along X axis, say with an acceleration a ⇒mg sin 0 = = ma

• The ball does not move perpendicular to the plane. It is in equilibrium along y'

• axis

• ⇒mg cos OR = 0

Answer 2

Answer:$= {sec}^{2} θ$

Solution:

The ball moves along x−axis with an acceleration, say a, whereas it is equilibrium along y axis.

$⇒∑ F_{x} = 0⇒R \: sin \: θ=ma$

$and \: ∑ F_{y}=0⇒R \: cos \: θ=mg$

⇒=mg sec θ (1)

When F= 0, Next, the ball moves down the plane, along X axis, say with an acceleration a

⇒mg sin θ = ma

The ball does not move perpendicular to the plane. It is in equilibrium along y ′−axis

⇒mg cosθR = 0

⇒R = mg cos θ

Dividing (1) by (2)⇒$\frac{R}{R′} = {sec}^{2} θ$