A small block of weight 100n is placed to an inclined plane which makes an angle of 30 with the horizontal the components of the weight perpendicular and parallel to the plane are respectively
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Refer to the attachment .
Here mg = 100N
∅ = 30°
The component of force parallel to the plain will be mgsin∅ or 100×1/2 => 50N
The component perpendicular to the plane is Mgcos∅ => 100×√3/2 = 50√3 .
Here mg = 100N
∅ = 30°
The component of force parallel to the plain will be mgsin∅ or 100×1/2 => 50N
The component perpendicular to the plane is Mgcos∅ => 100×√3/2 = 50√3 .
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