A heavy piece of machinery is raised by sliding it a distance of d-10m along a plank oriented at an angle- - 60- as shown in figure- How far has it moved vertically and horizontally-

Using simple trigonometry x component = 10 cos 60^∘ = 10 × 1/2 = 5m y component = 10 sin 60^∘ = 10 × √(3)/2 = 8.66 m ...

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Standard XIII

Physics

Vector Component

A heavy piece...

Question

A heavy piece of machinery is raised by sliding it a distance of d=10m along a plank oriented at an angle, $θ = 60^{\circ}$ as shown in figure. How far has it moved vertically and horizontally?

x- 8.66m, y- 5m

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x - 5m, y- 9m

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x-5m, y-8.66m

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x-4.5m, y-8.66m

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Solution

The correct option is C
x-5m, y-8.66m

Using simple trigonometry

$x c o m p o n e n t = 10 c o s 60^{\circ} = 10 \times \frac{1}{2} = 5 m$

$y c o m p o n e n t = 10 s i n 60^{\circ} = 10 \times \frac{\sqrt{3}}{2} = 8.66 m$

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A heavy piece of machinery is raised by sliding it a distance of d=10m along a plank oriented at an angle, θ=60∘ as shown in figure. How far has it moved vertically and horizontally?

The correct option is C x-5m, y-8.66m Using simple trigonometry xcomponent=10cos60∘=10×12=5m ycomponent=10sin60∘=10×√32=8.66m

A heavy piece of machinery is raised by sliding it a distance of d=10m along a plank oriented at an angle, $θ = 60^{\circ}$ as shown in figure. How far has it moved vertically and horizontally?

The correct option is C
x-5m, y-8.66m

Using simple trigonometry

$x c o m p o n e n t = 10 c o s 60^{\circ} = 10 \times \frac{1}{2} = 5 m$

$y c o m p o n e n t = 10 s i n 60^{\circ} = 10 \times \frac{\sqrt{3}}{2} = 8.66 m$