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3 years ago

How does the use of simple machines affect force and distance when work is done?

Physics

1 answer:

bazaltina [42]3 years ago

8 0

Answer:

Explanation:

Machines simply make work easier to do. They increase the amount of force exerted on a body and also the distance through which the force is applied. Also, they can also change the direction through which force on them is applied in order to produce much more work.

Work done = force x distance

The input force in a machine is attenuated to yield even more force. This is the purpose of designing a simple machine. When the force increases, more work would be produce with our little effort applied on the body.

Work done is a function of the force applied on a body and the distance through which it moves.

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You throw a rock horizontally out of a 7th story window. You time that it takes 3.7 seconds to hit the ground, and measure that

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Since we are only looking at the vertical height, we can use the free fall equation to find the height:
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3 years ago

A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten

Lyrx [107]

Answer:

a) the maximum shear stress τ $_{max}$ the bar is 16T $_{max}$ /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

τ $_{max}$ = 16T $_{max}$ /πd³

Therefore, the maximum shear stress τ $_{max}$ the bar is 16T $_{max}$ /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta ( $\theta$ ) be the angle of twist

polar moment of inertia $I_p}$ = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d $\theta$ = T(x)dx / G $I_{p}$

d $\theta$ = tx.dx / G(πd⁴/32)

d $\theta$ = 32tx.dx / πGd⁴

so total angle of twist $\theta$ will be;

$\theta$ = $\int\limits^L_0 \, d\theta$

$\theta$ = $\int\limits^L_0 \,$ 32tx.dx / πGd⁴

$\theta$ = 32t / πGd⁴ $\int\limits^L_0 \, xdx$

$\theta$ = 32t / πGd⁴ [ L²/2]

$\theta$ = 16tL² / πGd⁴

Therefore, the angle of twist between the ends of the bar is 16tL² / πGd⁴

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Learn more about graphite:brainly.com/question/11095487

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