two conditions. break the ropes into verical and horizontal componednts.
Horizontal:
Fl*sinthetaL-FR*sinThetaR=0 (left and right)
Vertical
Fl*cosThetal+Fr*cosTehtaR=40 (sum of forces equal weight).
well, your angles makes it easy, all the trig functions are .707
so immediately, from the horizontal equation, we have Fr=Fl in magnitude
and the vertical
fl=fr=40/.707
A 40-N box is suspended in the ceilingusing two ropes . Each ropes makes an angle of 45° with the vertical . How much is the tension in each rope?
2 answers
T1*sin(90+45) + T2*sin45 = 40.
0.707T1 + 0.707T2 = 40.
T1 = T2. Replace T2 with T1:
0.707T1 + 0.707T1 = 40.
T1 = 28.3 N. = T2.
0.707T1 + 0.707T2 = 40.
T1 = T2. Replace T2 with T1:
0.707T1 + 0.707T1 = 40.
T1 = 28.3 N. = T2.
0 answers