Static Equilibrium Rope Problem

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To maintain static equilibrium at the knot where three ropes are tied, the forces exerted by the first two friends must be balanced by the force on the third rope. Friend One pulls with 3.0 N at 0 degrees, while Friend Two pulls with 5.0 N at 120 degrees. A force diagram is essential to visualize the vectors and calculate the necessary force and direction for the third rope. The solution requires applying vector addition and equilibrium equations to determine the magnitude and angle of the third force. Properly analyzing these forces will ensure the knot remains stationary.
Miss Figgy
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Homework Statement



Three ropes are tied together. One of your friends pulls on a rope with 3.0 units of force and another pulls on a second rope with 5.0 units of force. Friend Two is 120 degrees to the left of Friend One (where Friend One's vector is on the positive x-axis).

Homework Equations


How hard must you pull on the third rope to keep the knot from moving? And In what direction must you pull on the third rope to keep the knot from moving (counterclockwise from the left direction)?


The Attempt at a Solution


T(1)=3 N
T(2)=5 N
T(3)=? N

T(1)= 0 degrees
T(2)= 120 degrees
T(3)= ? degrees

HELP!
 
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Hi Miss Figgy,

Miss Figgy said:

Homework Statement



Three ropes are tied together. One of your friends pulls on a rope with 3.0 units of force and another pulls on a second rope with 5.0 units of force. Friend Two is 120 degrees to the left of Friend One (where Friend One's vector is on the positive x-axis).

Homework Equations


How hard must you pull on the third rope to keep the knot from moving? And In what direction must you pull on the third rope to keep the knot from moving (counterclockwise from the left direction)?


The Attempt at a Solution


T(1)=3 N
T(2)=5 N
T(3)=? N

T(1)= 0 degrees
T(2)= 120 degrees
T(3)= ? degrees

HELP!

Have you drawn a force diagram for this problem? What equations do you get from it?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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