Forces, i have already attempted the problem.?

Question

The steel I-beam is basically a crane with a triangle it is liftling. the triangle is isosceles with two 70 degree angles and one 40 degree angle. it has a weight of 9.30 kN and is being lifted at a constant velocity. What is the tension in each cable attached to its ends?


i know that the velocity is constant so a=0, and the sum of the forces =0. and i know mg = 9300 N, not 9300*9.8 m/s^2. You do not multiply by 9.8, since you are given the weight, not the mass.

so....

t1~tention 1
t2~tention 2
in the y dirrection: t1 sin theda1 + t2 sin theda2=0
in the x dirrection, t1 cos theda1 + t2 cos theda2=0

i understand that if you make a triangle, t1 is the hypotenuse, t1y=1/2 fg, and t1x would be one the bottom.

so t1y=1/2mg/sin70

that is as far as i could get with this problem. see, i really did try to do it. i just can't finish it. please, PLEASE help me. i really am trying!

Answer 1

SInce you know that the forces add up to o, you understand trig, then, just keep adding the sides. The way you expressed the problem makes it hard to help you, but, these are not hard. Rest and do it again.