A steel beam (with linear mass density of 6.88 kg/m) has a total length of 10.0m and is suspended by two cables such that the beam hangs horizontally. Cable A is attached 1.00 m from the left end of the beam and cable B is attached 3.00 m from the right end of the beam. A 75.0 kg man is standing on the beam a distance of 2.5 m from the left end and a 3.5 kg cat is sitting, with a wry smile, 1.00 m from the right end of the beam. Assuming the cables are massless, what is the tension in Cable A?_________N What is the tension in cable B? _________N Some time later, after the man has kicked the cat off of the beam, the man is standing at the right edge of the beam. Should he be worried about the stability of this situation? explain.___________________________________________

Answer 1

In anticipation of part B I am going to take moments about the cable A attachment
I will do it all in Kg instead of Newtons because g will cancel except at the end.
m positive down
cable A x = 0 m = A
Man x = 1.5 m = 75
beam x = 4 m = 68.8
cable B x = 6 m = B
cat x = 8 m = 3.5

total m = 0 = A + B + 75 + 68.8 + 3.5
so
A + B = -147.3
now moments or torques about A
0 = 75*1.5+68.8*4+B*6+3.5*8
-6B = 415.7
B = - 69.3 Kg * 9.8 = 679 Newtons up tension
then
A -69.3 = -147.3
A = -78 * 9.8 = 764 Newtons up tension

Now with no cat and the man at the right
A + B = -143.8
0 =75*9 +68.8*4+B*6
B = -158.4
Oh my, that means that A will have to be positive (force down) to make the A+B=-143.3
The cable A will be under negative tension. Cables do not so well under negative tension, they collapse and the left end goes up and the man falls off.

Answer 2

Could you explain the meaning of your symbols you use. for example the "m positive down" Ok i know what you mean, but this is extremely confusing if you don t state what each symbol means beforehand.
Cheers