Question

A mass M=161 kg is suspended from the end of a uniform boom as shown. The boom (mass=75.0 kg, length=3.60 m) is at an angle θ=65.0 deg from the vertical, and is supported at its mid-point by a horizontal cable and by a pivot at its base. Calculate the tension in the horizontal cable.

Answer 1

Let the tension in the cable be T. Calculate torque about the base of the beam. If length of the beam be 2l and the mass of it be m then, left hand torque = right hand torque.
Left hand torque = \[T \times l \cos \theta\]
Right hand torque = \[mg \times l \sin \theta + Mg \times 2l \sin \theta\]
So, \[ T =(m + 2M)g \tan \theta\]
Putting the numerical values, take \[g = 10 m/s^2 \] \[T = (75 + 2 \times 161) \times 10 \times tan \theta \]
\[T = 3970\times tan \theta \ N\]
\[T = 5836 N \]

Answer 2

It says that it's wrong, would you know why? I really appreciate your time!!

Answer 3

Is the numerical value of the result
wrong ?

Answer 4

yeah

Answer 5

That may be. Put the correct values in the result \[T = (m + 2M)g \tan \theta\] If still it is wrong, let me know. Or tell me the error.

Answer 6

Okay, so what would go into the tan then?

Answer 7

Put \[\tan 65 ^{0} = 1.47\]

Answer 8

What is the unit of the answer ?

Answer 9

It is in Newtons

Answer 10

Oh I got it! I think we just calculated it wrong. The answer is 8343 N...Thank you so much!

Answer 11

How come ? Please, send me your solution.

Answer 12

(75+(2*161))*9.8*tan(65)=8343N

Answer 13

So I put wrong value of tan65. Thanks. :)

Answer 14

No, thank you! :)