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3 years ago

what output force is generated when an input force of 630 n is applied to a machine with a mechanical advantage of 3

Physics

2 answers:

Natali5045456 [20]3 years ago

8 0

Answer : Output force is 1890 N.

Explanation :

Given that,

Input force is 630 N

Mechanical advantage is 3

Mechanical advantage of a machine is defined as the ratio of output force and the input force.

Mathematically it can be written as :

$m=\dfrac{F_o}{F_i}$

Where,

$F_o$ is the output force

$F_i$ is the input force

So, $3=\dfrac{F_o}{630\ N}$

$F_o=1890\ N$

Hence, this is the required solution.

Marta_Voda [28]3 years ago

3 0

The mechanical advantage is the factor by which
the machine multiplies the input force.

If the MA is 3 and the input force is 630N, then
the output force is

(3) x (630N) = 1,890N

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So we can write:

$KE_i = PE_f$

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where

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v = 6.4 m/s is the initial speed of the sphere

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h is the maximum height reached

Solving for h, we find

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Calling $\theta$ the angle that the rope makes with the vertical, we can write

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$h=L(1-cos \theta)$

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The motion of the sphere is part of a circular motion. The forces acting along the centripetal direction are:

- The tension in the rope, T, inward

- The component of the weight along the radial direction, $mg cos \theta$ , outward

Their resultant must be equal to the centripetal force, so we can write:

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$T-mg cos \theta=0$

So we can find the tension:

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We can solve this part by applying again the law of conservation of energy.

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$KE_i = KE_f + PE_f\\\frac{1}{2}mv^2 = \frac{1}{2}mv'^2 + mgh'$

where:

$KE_i$ is the initial kinetic energy

$KE_f$ is the kinetic energy at 1 m

$PE_f$ is the final potential energy

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v' is the speed at a height of 1 m

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And solving for v', we find:

$v'=\sqrt{v^2-2gh'}=\sqrt{6.4^2-2(9.8)(1)}=4.6 m/s$

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$T-mg cos \theta = m\frac{v^2}{r}$

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$mg cos \theta$ is the component of the weight in the radial direction

$m\frac{v^2}{r}$ is the centripetal force

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v = 4.6 m/s is the speed of the sphere

$cos \theta$ can be rewritten as (see part c)

$cos \theta = 1-\frac{h'}{L}$

where in this case,

h' = 1 m

L = 4 m

And $r=L=4 m$ is the radius of the circle

Substituting and solving for T, we find:

$T=mg cos \theta + m\frac{v^2}{r}=mg(1-\frac{h'}{L})+m\frac{v^2}{L}=\\=(3)(9.8)(1-\frac{1}{4})+(3)\frac{4.6^2}{4}=37.9 N$

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