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A 90 Kg Painter Is Standing On 4 Kg Board, Which Is 12 M Long And Suspended On Each End By A Rope. The Painter Is Standing 3 M From The Left Side, Closer To The First Rope. Q1) Find The Tension In The First Rope. Q2) Find The Tension In The Second Rope. Q3) The Painter Climbs Down From The Board, Just Before The Second Rope Suddenly Snaps. Find The instantaneous angular acceleration of the board

Respuesta :

The board's instantaneous angular acceleration is 1.225 rad/sec, with the first rope's tension being 681.1 N and the second rope's tension being 240.1 N.

A force along a medium's length is known as tension, particularly a force carried by a flexible medium like a rope or cable. The equivalent of linear acceleration in rotation is called angular acceleration. Angular acceleration is the rate of change of angular velocity, whereas linear acceleration describes the rate of change of linear velocity.

T2*12 - 90g*3 - 4g*6 =0

T2 = 240.1N for the Second Rope's Tension

T1+T2 = 921.2

T1 = 681.1N for the first rope's tension

The board's instantaneous angular acceleration is given by T = I(alpha), which is equal to T = 4(12)^2/3 6*9.8*3/144 = alpha.

The board's instantaneous angular acceleration is equal to 1.225 rad/sec.

Learn more about angular acceleration here

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