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Svetach [21]
3 years ago
3

A museum curator moves artifacts into place on many different display surfaces. Consider the following table giving approximate

values for coefficients of friction: – schomer – (HHSPhy4F18) 1 Materials μs μk steel on steel 0.74 0.57 aluminum on steel 0.61 0.47 rubber on dry concrete 1.0 0.8 rubber on wet concrete – 0.5 wood on wood 0.4 0.2 glass on glass 0.9 0.4 waxed wood on wet snow 0.14 0.1 waxed wood on dry snow – 0.04 metal on metal (lubricated) 0.15 0.06 ice on ice 0.1 0.03 Teflon on Teflon 0.04 0.04 synovial joints in humans 0.01 0.003 What is Fs,max for moving a 167 kg alu- minum sculpture across a horizontal steel platform? The acceleration of gravity is 9.81 m/s2 . Answer in units of N.
Physics
1 answer:
dimaraw [331]3 years ago
3 0

you must give a force for just moving  

Fmax=151*0,61*9,81= 904 N <--------------  

and  

F=151*0.47*9.81= 646 N ... to keep in motion

hope this helps :)

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Answer:

<u>6.87 ft/s</u> is the rate at which the top of ladder slides down.

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Given:

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h^2+b^2=L^2\\h^2+b^2=20^2\\h^2+b^2=400----1

Differentiating the above equation with respect to time, 't'. This gives,

\frac{d}{dt}(h^2+b^2)=\frac{d}{dt}(400)\\\\\frac{d}{dt}(h^2)+\frac{d}{dt}(b^2)=0\\\\2h\frac{dh}{dt}+2b\frac{db}{dt}=0\\\\h\frac{dh}{dt}+b\frac{db}{dt}=0--------2

In the above equation the term \frac{dh}{dt} is the rate at which top of ladder slides down and \frac{db}{dt} is the rate at which bottom of ladder slides away.

Now, as per question, h=8\ ft, \frac{db}{dt}=3\ ft/s

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8^2+b^2=400\\64+b^2=400\\b^2=400-64\\b^2=336\\b=\sqrt{336}=18.33\ ft

Now, plug in all the given values in equation (2) and solve for \frac{dh}{dt}

8\times \frac{dh}{dt}+18.33\times 3=0\\8\times \frac{dh}{dt}+54.99=0\\8\times \frac{dh}{dt}=-54.99\\ \frac{dh}{dt}=-\frac{54.99}{8}=-6.87\ ft/s

Therefore, the rate at which the top of ladder slide down is 6.87 ft/s. The negative sign implies that the height is reducing with time which is true because it is sliding down.

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