Physics, asked by pepitocena9, 7 months ago

An object 8.9 N in weight is released from a height H = 30.1 m so that it can rotate in the loop of radius r = 4.8 m and continue on its way to a rough horizontal section of Length L = 36.2 m, to stop at D. Only the CD section is rough. h = 1 m a) the speed of the object at the highest part of the loop (point B). b) the kinetic friction coefficient μC of the section CD. pls help Or my teacher will think that I'm a human trash for the rest of my days

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Answered by HariesRam

The formula to calculate the coefficient of friction is μ = f÷N. The friction force, f, always acts in the opposite direction of the intended or actual motion, but only parallel to the surface.

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Answered by SwaggerGabru

$\huge\red{\underline{{\boxed{\textbf{QUESTION}}}}}$

An object 8.9 N in weight is released from a height H = 30.1 m so that it can rotate in the loop of radius r = 4.8 m and continue on its way to a rough horizontal section of Length L = 36.2 m, to stop at D. Only the CD section is rough. h = 1 m a) the speed of the object at the highest part of the loop (point B). b) the kinetic friction coefficient μC of the section CD.

$\huge\red{\underline{{\boxed{\textbf{ANSWER}}}}}$

We know that according to polygon law of vector addition, the resultant of these six vectors is zero.

Here A = B = C =D=E=F (magnitude) So, Rx = A cosi + A cos /3 + A cos 2r/3 + A cos 3r/3 + A cos 4/4 + A cos 5x/5=0 As resultant is zero. X component of resultant R = 0] = cos 8 + cos /3 + cos 2n/3 + cos 33 + cos 4x/3 + cos 5/3 = 0 Note: Similarly it can be proved that sin 0+ sin /3 + sin 2/3 + sin 3/3 + sin 4/3 + sin 5a/3 = 0

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