What is the Solution to a Tension Problem using Trigonometry?

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To solve the tension problem using trigonometry, the goal is to find the smallest force P that makes the resultant R of two forces at point A vertical. The horizontal component of force Q must be canceled by P, leading to the calculation of P as 45.9 using the horizontal component of Q (80sin(35)). The angle alpha is determined to be a right angle, which is essential for minimizing P. Subsequently, the Law of Sines can be applied to find the third angle, and the Law of Cosines is used to calculate the magnitude of R. This approach effectively utilizes trigonometric principles to address the tension problem.
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Homework Statement


Two cables are attached to a sign at point A to steady the sign as it is being lowered. Using Trigonometry, determine a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied at A is vertical, b) the corresponding magnitude of R


Homework Equations


Use Law of cosines (R^2 = Q^2 + P^2 - 2QPcos(theta))
and Law of sines P/sin(opp. angle) = Q/sin(opp. angle)

The Attempt at a Solution


diagram is shown in attachment. I only have one magnitude and one angle and I'm supposed to get an angle alpha so that P is the min.
 

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I have no idea where ure stuck at but ill drop a small hint:

The horizontal component of force Q must cancel for the resultant to be vertical.
 
Right, thanks. it's the small things that get me. taking the horz component of Q (80sin(35) = P = 45.9) gives the magnitude of p. Then, using law of sines to get alpha, I can get the third angle and then use law of cosines to get mag of R
 
jonnyboy said:
Right, thanks. it's the small things that get me. taking the horz component of Q (80sin(35) = P = 45.9) gives the magnitude of p. Then, using law of sines to get alpha, I can get the third angle and then use law of cosines to get mag of R

if you have already noticed if you went ahead with the above, alpha should be a right angle since the smallest force you need with p is a horizontal force to cancel out with the horizontal component of Q.
 
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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