Answer:
Step-by-step explanation:
Let A and B represent the angles that PA and PB make with the horizontal, respectively. Let Ta and Tb represent the tensions in PA and PB, respectively. Then for equilibrium, we have ...
Ta·cos(A) - Tb·cos(B) = 0 . . . . . horizontal equilibrium
Ta·sin(A) +Tb·sin(B) = T . . . . . . where T is the downward force exerted by P
__
The general solution of this pair of equations is ...
Ta = T·cos(B)/(sin(B)cos(A) +sin(A)cos(B)) . . . by Cramer's rule
which simplifies to
Ta = T·cos(B)/sin(A+B)
Tb = T·cos(A)/sin(A+B) . . . . by symmetry or Cramer's rule
If we call the position of the right angle point X, then the Pythagorean theorem tells us AX = 30 (3-4-5 triangle) and BX = 96 (5-12-13 triangle). Then the cable tensions are ...
Ta = T·(12/13)/(12/13·4/5 +3/5·5/13) = T·(20/21)
Ta = (2.1 kg)(9.8 m/s²)(20/21) = 19.6 N
Tb = T·(3/5)/(63/65) = T·(13/21)
Tb = (2.1 kg)(9.8 m/s²)(13/21) = 12.74 N
These tension values can be rounded to Ta = 20 N, Tb = 13 N.
_____
The weight of P is ...
F = Ma = (2.1 kg)(9.8 m/s²) = 20.58 N
