(adsbygoogle = window.adsbygoogle || []).push({}); 1. An 85-kg man plans to tow a 109000-kg

airplane along a runway by pulling horizontally on a cable attached

to it. Suppose that he instead attempts the feat by pulling the cable at

an angle of 9.00 above the horizontal. The coefficient of static friction

between his shoes and the runway is 0.77. What is the greatest acceleration

the man can give the airplane? Assume that the airplane is on

wheels that turn without any frictional resistance.

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MY ATTEMPT AT THE SOLUTION

Forces on Plane

Fn = Normal force

Weight = (mass of plane)(gravity) = 109000(9.8) = 1068200 N

Tx = x-component of tension in rope = Tcos(9)

Ty = y-component of tension in rope = Tsin(9)

Forces on Man

Fn = Normal force

Weight = (mass of man)(gravity) = 85(9.8) = 833 N

Fk = frictional force = (weight)(coefficient of friction) = (833)(0.77) = 641.41 N

Tx = x-component of tension in rope = -Tcos(9)

Ty = y-component of tension in rope = -Tsin(9)

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net equations for the plane

Fx = Tcos(9) = (Mass of plane)(x-component of acceleration)

Fy = Tsin(9) = (Mass of plane)(y-component of acceleartion)*********

net equations for the man

Fx = Fk - Tcos(9) = (Mass of man)(x-component of acceleration)

Fy = -Tsin(9) = (Mass of man)(y-component of acceleration)*********

From here, i think it is plug and chug to find tension. After I find tension, I can find both the x and y components of acceleration. I can then use Pythagorean theorem to find acceleration from the x and y components.

My question is, have I written my net equations correctly???? I think something is wrong because when I plug in T to the two starred equations above, I get two different answers for the y-component of acceleration...help??? please?

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# Friction, tension, and acceleration combination problem

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