Hard Physics Question?

Question

A 58.2 kg sign is suspended by two wires, as the drawing shows. Find the tension in wire 1 and in wire 2.
http://www.webassign.net/CJ/4-58alt.gif

Answer 1

Forces are F1 ( the right ) and F2.

Summation of Fx=0
Summation of Fy=0


Summation of Fx= -F1cos(43)+ F2cos(55)=0
Summation of Fy= F1sin(43)+ F2sin(55) - mg=0

Two equation and two unknowns

-F1 cos(43) sin(43)+ F2cos(55)sin(43) =0
F1 sin(43) cos(43)+ F2 sin(55) cos( 43)- mg cos(43)=0
adding we get
F2(cos(55)sin(43) + sin(55) cos( 43) ) - mg cos(43)=0
F2= mg cos (43)/(cos(55)sin(43) + sin(55) cos( 43) )

then since F1 = F2cos(55)/cos(43)

This a done deal!

Have fun

Answer 2

Since the sign is stationary, the force vectors have to total zero. This works independently in the x and y directions.

The tension in wire 1 makes a angle of 180-43=137 degrees with the positive x direction.
The tension in wire 2 makes an angle of 55 with the positive x direction.
The force of gravity on the sign makes an angle of 270 degrees with the positive x direction and has a magnitude of 58.2 kg * 9.8 m/sec^2= 570.36 N

x direction:
t1 * cos(137) + t2 * cos(55) + 570.36 N * cos(270) = 0

y direction:
t1 * sin(137) + t2 * sin(55) + 570.36 N * sin(270) = 0

cos(137) = -cos(43); sin(137) = sin(43)
cos(270) = 0; sin(270) = -1

so...
x: t1 * cos(43) = t2 * cos(55) ==> t1 = t2 cos(55)/cos(43)
y: t1 * sin(43) + t2 * sin(55) = 570.36 N

Substitute the expression for t1 into the y equation and find a value for t2. Then substitute that value back into the x equation to find t1.

t2 cos(55)/cos(43) * sin(43) + t2 * sin(55) = 570.36 N
t2 (cos(55)/cos(43) + sin(55)) = 570.36 N
t2 = 570.36 N / ((cos(55)/cos(43) + sin(55))
Then...
t1 = t2 cos(55)/cos(43)

(Or you could use a series of matrix operations to solve the simultaneous equations. It works out the same.)

Answer 3

i guess the person above me has it ,but all i hv to say is how much i hate to do webassigns ...... :)