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2 years ago

In order to lift a lighter object on the other side, a boy placed 155 N of

Engineering

1 answer:

const2013 [10]2 years ago

4 0

Answer:

0.58 m

Explanation:

In order to find the vertical distance the load lifts, we first need to find the magnitude of the resistance force. So, Rd = Fd' where R = resistance force = ?, d = resistance distance = 3.5 m, F = effort force = 155 N and d' = effort distance = 1.5 m.

So R = Fd'/d = 155 N × 1.5 m/3.5 m = 232.5 Nm/3.5 m = 66.43 N

Now, the vertical work done by the effort = vertical work done by load

Fy = Ry' where y = vertical distance moved by effort = 0.25 m and y' = vertical distance moved by resistance force

So, Fy = Ry'

y' = Fy/R

= 155 N 0.25 m/66.43 N

= 38.75 Nm/66.43 N

= 0.58 m

So, the vertical distance the lighter object moves is y' = 0.58 m

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$\theta_1=15^o\\\theta_2=75^o$

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<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

$V_x=V_{ox}=V_ocos\theta$

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$x=V_{o}cos\theta t$

$\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}$

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$\displaystyle t_f=\frac{2V_osin\theta}{g}$

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$\displaystyle R=\frac{V_o^2sin2\theta}{g}$

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$\displaystyle sin2\theta=\frac{R.g}{V_o^2}$

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