Asked by samson

A 50kg boy suspends himself from a point on a rope tied horizontally between two vertical poles. The two segments of the rope are then inclined at angles 30 degrees and 60 degrees respectively to the horizontal.The tensions in the segments of the rope in
Answered by MathMate
The few problems you have posted at the same time indicates that you are in the process of learning how to draw a free-body diagram (fbd).

Follow your instructor's directions to draw a fbd and identify the triangle to be solved.


You should end up with a right triangle in which the sides represent the three forces, weight of boy (hypotenuse, 50g), and tension of each of the strings. The angles of the triangle are 30-60-90 so it would be easy to solve.
There is no graphic capabilities here, so I could not have illustrated here.
Answered by Henry
M*g = 50 * 9.8 = 490 N. = Wt. of boy.

T1*Cos(180-30) + T2*Cos60 = 0.
-0.866T1 + 0.5T2 = 0.
T2 = 1.732T1.

T1*sin(180-30) + T2*sin60 = 490
0.5T1 + 0.866T2 = 490.
Replace T2 with 1.732T1:
0.5T1 + 0.866*1.732T1 = 490.
2T1 = 490.
T1 = 245 N.
T2 = 1.732 * 245 = 424.3 N.

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