Physics Torque/Mechanics II Question

[yestomeds]

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Hello, this is my 1st time posting a study-type question here, so apologies if I'm not doing it correctly or if I'm breaking any rules, or etc.

Ok, so my question is from an online TPR section.
What I need help with is a way of using equations (formulas, numbers) etc. to arrive at the solution. Because just reading their verbal explanation isn't doing it for me (it sounds great, must be the correct answer, etc. but I'm not truly "getting" it).

Ok, so if you have an alternate way of understanding it, or if you have a way of using the equations/formulas relevant to the chapter, please help me! Thank you 🙂.




QUESTION: How far from the heavier end must the fulcrum of a massless 5-m seesaw be if an 800-N man on one side is to balance his 200-N daughter at the other end?

ANSWER: To balance the torques due to the weight of the man and of his daughter, the fulcrum must be placed 4 times farther from the daughter than from her father, because he weighs 4 times more than she does. Since the total length of the seesaw is 5 m, the fulcrum must be placed 4 m from the daughter and 1 m from the man (that is, from the heavier end).

 

[DrknoSDN]

For balancing torque rotation you just use force times level arm length where the two distances are X and Y. Usually even with 2 variables its simple numbers that can be resolved without equations.

800N * Xmeters = 200N * Ymeters
Where: X+Y=5

You can almost guess and check because the numbers are easy. You plug in 1 for X giving 800N for his side, then plug in 4 for Y giving (200*4=) 800N for her side, and they are balanced.
He is 1 meter from fulcrum, and she is 4 meters from fulcrum.


You could also say:
800N * X = 200N * (5-X)

Here you just solve for X which is the fathers distance from the fulcrum.

 

[Teleologist]

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