Calculate Tension in String 1 for Hanging Sign Problem

[chaotixmonjuish]

I'm working on this hanging sign problem:

A sign hangs precariously from your prof's office door. Calculate the magnitude of the tension in string 1, if θ1 = 27.42°, θ2 = 67.35°, and the mass of the sign is 9.9 kg.

I was wondering if I could put 9.9*9.8 as a leg in the middle of the triangle to calculate the tension on the two strings. The only given hint is to calculate components.

 

[stewartcs]

Draw a FBD then use Newton's second law, Fnet = Ma

T1 + T2 + Fg = Ma

The sign is in static equilibrium so Ma = 0, yielding,

T1 + T2 + Fg = 0

Find the x components and the y components using the given angles (noting their referenced axis) and then solve the two resulting simultaneous equations.

CS

 

[chaotixmonjuish]

could i use the force the sign is exerting as a leg in order to aid with component calculations?

 

[chaotixmonjuish]

I guess what I'm trying to say, am I already give a Y component 97.02 N, and can I use that to calculate the remaining ones

 

[stewartcs]

No. With only the Y component of force you do not have enough information to solve the problem. You must use Newton's second law and then break the problem down into componets in order to find the tension vector.

The technique I posted is the proper way to solve the problem so you will conceptually understand what is going on with all of the forces.

 

[chaotixmonjuish]

can i just make sure I have this right
t
T1+T2-Mg=Ma

T1+T2-97.02=0

That's what I got from my relatively simple diagram. I'm just not sure how to calculate components without a length.

 

[stewartcs]

T1 + T2 - Mg = 0 is correct. This is your equilibrium equation.

The components depend on the referenced axis and the direction of the angle. Is θ1 = 27.42° from the positive x-axis or negative x-axis (look at your drawing). What about θ2 = 67.35° referenced position (they should be opposite given your problem but maybe not)?

If θ1 is a negative angle from the negative x-axis and θ2 from the positive you have:

x components: -T1cos(θ1) + T2cos(θ2) - 0 = 0

y components: T1sin(θ1) + T2sin(θ2) - Mg = 0

Solve those two equations simultaneously and then substitute back into the equilibrium equation.

 

[chaotixmonjuish]

I know this is probably going to yield me a blaring duh, but i just solve the equation for T1.

 

[chaotixmonjuish]

okay, so i forgot to mention that T2 is to the left of the Y

my two equations are:

X=-T2*cos(theta)+T1*cos(theta)=0
Y= T1*sin(theta)+T2*sin(theta)-97.02=0

I tried solving for X
T2*cos(theta)=T1*cos(theta)
T2=T1*cos(theta)/cos(theta)
T2=T1

It worries, me, I don't think that should be right.

 

[chaotixmonjuish]

I got 66.4289 N for the tension in string 1

is this right?

 

[chaotixmonjuish]

Anyone?

 

[stewartcs]

okay, so i forgot to mention that T2 is to the left of the Y

my two equations are:

X=-T2*cos(theta)+T1*cos(theta)=0
Y= T1*sin(theta)+T2*sin(theta)-97.02=0

I tried solving for X
T2*cos(theta)=T1*cos(theta)
T2=T1*cos(theta)/cos(theta)
T2=T1

It worries, me, I don't think that should be right.

No. The angles (theta 1 and theta 2 are different) so T1 is not equal to T2. If they were the same then you would be correct.

 

[chaotixmonjuish]

do the cos not cancel out

 

[stewartcs]

No. The angles are different so the cosines do not cancel. If the angles were the same they would, but they are not.

 

[chaotixmonjuish]

Is the answer 187.011 N? I just got that by turning the whole thing into a triangle.

 

[stewartcs]

No.

Did you solve these two equations simultaneously?

x components: -T1cos(θ1) + T2cos(θ2) - 0 = 0

y components: T1sin(θ1) + T2sin(θ2) - Mg = 0


Solve the x component equation for T2:

-T1cos(θ1) + T2cos(θ2) - 0 = 0

T2cos(θ2) = T1cos(θ1)

T2 = T1(cos(θ1)/cos(θ2))


Now substitue T2 into the y component equation:

T1sin(θ1) + T2sin(θ2) - Mg = 0

T1sin(θ1) + T1(cos(θ1)/cos(θ2))*sin(θ2) = Mg

T1(sin(θ1) + (cos(θ1)/cos(θ2))*sin(θ2)) = Mg

T1 = Mg/(sin(θ1) + (cos(θ1)/cos(θ2))*sin(θ2))

Now you have the two equations (assuming the referenced angles were the same as I posted earlier). Solve this one for a numeric value of T1 since you have all the missing information for it. Then plug that answer into the T2 = T1(cos(θ1)/cos(θ2)) to find the other tension.