A horizontal uniform bar of mass 2.4 kg and length 3.0 m is hung horizontally on

Question

A horizontal uniform bar of mass 2.4 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the end of the bar, and string 2 is attached a distance 0.7m from the other end. A monkey of mass 1.2 kg walks from one end of the bar to the other. Find the tension in string 1 at the moment that the monkey is halfway between the ends of the bar.

Find T1, the magnitude of the force of tension in string 1, at the moment that the monkey is halfway between the ends of the bar.

Explanation / Answer

By newtons 2nd law, we have F= ma

a= 0

Force down = 2.4g +1.2g =3.6g

Force up = T1 + T2

T1 +T2 - 3.6g = 0

=> T1+T2 =3.6g ------------ (1)

Now the equation of torque we have

= I

=0

=> the is cancels out

around the centre of rod is T1*1.5 - T2*(1.5-0.7) =0

=> T2/T1 = 1.87

   => T2 = 1.87T1

now using this in equation 1 we get

T1 +1.87T1 = 3.6g

=> T1 = 1.25g

=> T1 = 12.27 N    (since g = 9.8 m/s^2)

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