+185 a 450 pound weight is suspended by two ropes. if one makes an angle of 24 degrees with the vertical and the other an angle of 32 degrees, what is the tension in each rope?
+36916 Statics: Since the object is hanging in equalibrium, all of the vertical forces have to balance(i.e. sum= o) and all of the horizontal forces have to balance.
Vertical forces:
32rope = tcos32
24rope= t'tcos24
weight = -450
SO they must balance: tcos32 + t'cos24 - 450 =0
tcos32 +t'cos24 = 450 (FIRST equation)
Horizontal forces must cancel out too: (the horizontal force in 32rope must equal horizontal force in 24rope)
32rope = t sin 32
24rope = t' sin 24 These forces are EQUAL (if they were not EQUAL something would MOVE)
tsin32 = t'sin24 we can solve this for t and substitute into the FIRST eqaution for VERTICAL forces to solve
t = t' sin 24/sin32 (SECOND equation)
Substitute: (t' sin24/sin32) * cos32 + t'cos24 = 450
t' (.6509 + .9135) = 450 t' = 287.64 lbs (tension in 24 degree rope)
SECOND equation t = t' sin 24/sin32 t = 220.77 lbs (tension in 34 degree rope)
+36916 Statics: Since the object is hanging in equalibrium, all of the vertical forces have to balance(i.e. sum= o) and all of the horizontal forces have to balance.
Vertical forces:
32rope = tcos32
24rope= t'tcos24
weight = -450
SO they must balance: tcos32 + t'cos24 - 450 =0
tcos32 +t'cos24 = 450 (FIRST equation)
Horizontal forces must cancel out too: (the horizontal force in 32rope must equal horizontal force in 24rope)
32rope = t sin 32
24rope = t' sin 24 These forces are EQUAL (if they were not EQUAL something would MOVE)
tsin32 = t'sin24 we can solve this for t and substitute into the FIRST eqaution for VERTICAL forces to solve
t = t' sin 24/sin32 (SECOND equation)
Substitute: (t' sin24/sin32) * cos32 + t'cos24 = 450
t' (.6509 + .9135) = 450 t' = 287.64 lbs (tension in 24 degree rope)
SECOND equation t = t' sin 24/sin32 t = 220.77 lbs (tension in 34 degree rope)
