Can you please show me all the steps in order to solve this problem.
Thanks
http://d.yimg.com/hd/answers/i/a3f83a466cc84308b085d7cf3cf62088_A.png?a=answers&mr=0&x=1382481888&s=2d72493d6f5ea03045244c9448198f63
Thanks
http://d.yimg.com/hd/answers/i/a3f83a466cc84308b085d7cf3cf62088_A.png?a=answers&mr=0&x=1382481888&s=2d72493d6f5ea03045244c9448198f63
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We are given :
m= 32.0 kg
g = 9.81m/s^2
We are looking for the Tension
We you look at the diagram you can see that the backpack isnt moving, it is in equilibrium , therefore the acceleration is 0. and we know that when the acceleration is 0 , the sum of all the forces in the system equals to zero.
For the x-components:
the sum of the F= 0
and Sum F = Tcos35 - T cos 35
recall the sum F =0 so 0 = Tcos35 - Tcos35 which doesnt really help so lets look at the y-components:
Sum F = 0
Sum F = 2Tcos35 - Fg ( we know that Fg = mg )
0= 2Tcos35 - mg
mg = 2T cos 35
mg / (2cos 35) = T
[(32)(9.81)] / (2cos35) = Tension = 273.37
thus, the tension of the rope is 273 N
m= 32.0 kg
g = 9.81m/s^2
We are looking for the Tension
We you look at the diagram you can see that the backpack isnt moving, it is in equilibrium , therefore the acceleration is 0. and we know that when the acceleration is 0 , the sum of all the forces in the system equals to zero.
For the x-components:
the sum of the F= 0
and Sum F = Tcos35 - T cos 35
recall the sum F =0 so 0 = Tcos35 - Tcos35 which doesnt really help so lets look at the y-components:
Sum F = 0
Sum F = 2Tcos35 - Fg ( we know that Fg = mg )
0= 2Tcos35 - mg
mg = 2T cos 35
mg / (2cos 35) = T
[(32)(9.81)] / (2cos35) = Tension = 273.37
thus, the tension of the rope is 273 N