Please help With Physics question, really desperate?

Question

Please help with this, i haven't gotten straight answers, ive gone halfway and im stuck, here's the question:


Consider a massless cord attached at both ends to hooks in a ceiling. A 20-kg mass is then suspended from the first cord by a second cord. The angles made by the cord with the ceiling are 40 degrees (left angle) and 25 degrees (right angle). Find the tension in each cord.


here are the two equations ive got

T1(.64) + T2(.42) = 0

and

T1(.77) + T2(.91) = 0

and im stuck

Answer 1

I DONT THINK YOU NEED THOSE EQUATIONS AT ALL.

YOU KNOW FG=M*-9.8,/S^2
= 20KG*-9.8M/S^2
=-196n

FN=-FG
FN=196n
YOU NEED TO DIVIDE FN BY TWO TO GET THE FY1 AND FY2
SO THEY EACH EQUAL 98n

T1=98n/(COS40)
= 127.9n
T2=98n/(COS25)
= 108.131n


HOPE THIS HELPS AND SORRY ABOUT THE CAPS

Answer 2

You have made a mistake.

The horizontal components are cosine parts and vertical components are sine parts.

There fore.

The left side force is P cos 40 = 0.77 P

The right side force is Q cos 25 = 0.91.

These two are equal but opposite.

0.77 P = 0.91 Q.

P = 1.18 Q. ------- (1).

The two vertical forces Psin 40 and Q sin 25 balances the weight 20 x 9.8 N.

P sin 40 + Q sin 25 = 196.

Substituting for P from equation (1)

1.181 Q sin 40 + Q sin 25 = 196 N.

0.76 Q + 0.42 Q = 196 N.

1.18 Q = 196 N

Q = 166 N and

P = 1.181 x 166 = 196 N.

P is almost equal to the weight of the suspended mass.

Answer 3

The weight hangs down vertically from the first chord. The sum of the vertical components of the chords' tension must equal m*g, the force acting on the weight. The tension forms the hypotenuse of a vector diagram, with the vertical component as on leg. That vertical leg is the side opposite the hypotenuse, so the sine of that angle is Fv/T, where Fv is the vertical component of force and T is the tension. The equation for both sides is the same, so the sum of the vertical components is

m*g = T1*sin(40º) + T2*sin(25º)

We have two unknowns here (T1 and T2) so we need another equation. That comes from the fact that the horizontal force components of the tension must be equal and opposite; the horizontal component are T*cos(angle), so

T1*cos(40º) = T2*cos(25º) (the chords pull in opposite directions, so the forces must be equal)

Solve the two like any two equations in two unknowns: T2 = T1*(cos(40º)/cos(25º), put into the first to get


m*g = T1*sin(40º) + [T1*(cos(40º)/cos(25º)]*sin(25º)

Multiply it out and solve for T1, then T2

Answer 4

sum the forces at the point where the two cords are connected

in the verticle direction: -20kg + Fleft cord x sin 40 + F right cord sin 25 = 0

in the horizontal direction: Fleft cord cos 40 = F right cord cos 25

substitute so that:

-20 kg + F left cord sin 40 + (F left cord cos 40/ cos 25) sin 25= 0