# Lever...1

Discussion in 'Algebra' started by nycmathguy, Feb 16, 2022.

1. ### nycmathguy

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Let w_1 = weight 1

Let w_2 = weight 2

Let d_1 = distance 1

Let d_2 = distance 2

I understand the equation for all lever problems to be (w_1)(d_1) = (w_2)(d_2).

1. John weighs 90 lbs and Jane weighs 60 lbs. They are both seating on a seesaw. If John is seated 10 feet away from Jane, how far should each be from the fulcrum of the seesaw?

I will use f for fulcrum.

John...........f..............Jane

Distance from John to f = x.
Distance from Jane to f = 10 - x

Equation

90x = 60(10 - x)

Yes?

2. John wants to move a 400 lb. rock with a 5 ft. 9 in. crowbar. He puts the fulcrum 9 inches from the rock. How much force must he use to move the rock?

Let 5 feet, 9 inches be 70.8 inches.

Rock...........f...........x

Let x = amount of force needed.

Distance from Rock to f = 9.
Distance from John to f = 70.8.

Equation

9(400) = 70.8x

Yes?

nycmathguy, Feb 16, 2022

2. ### Country Boy

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That "formula" comes from the fact that in problems like this
"torque" (also called "turning moment") must balance. If force F is applied at distance d from the pivot, the "torque" is Fd. (Strictly speaking F and d are both vector quantities. "Fd" here is not the scalar product, it is the "cross product", F x d, a third vector perpendicular to both F and d, the "axis of rotation".)

1. John weighs 90 lbs and Jane weighs 60 lbs. They are both seating on a seesaw. If John is seated 10 feet away from Jane, how far should each be from the fulcrum of the seesaw?

I would set up a "coordinate system" with John at 0 and Jane at 10. Let x be the position of the pivot be x. The distance from John to the pivot is x, in feet, so his torque is 90x "pound feet", counter clock wise. The distance from Jane to the pivot is 10- x so her torque is 60(10- x) clock wise. If they balance then 90x= 60(10- x). Yes, that is what you have. Good work! (I did then find the actual position of the pivot between John and Jane but since you don't want to know that I won't post it.)

2. John wants to move a 400 lb. rock with a 5 ft. 9 in. crowbar. He puts the fulcrum 9 inches from the rock. How much force must he use to move the rock?

Again, I would imagine a coordinate system, this time with 0 at the rock and 5 ft 9 in at the end of the crowbar (you say "Let 5 feet, 9 inches be 70.8 inches." You can't do that! 5 feet is 5(12)= 60 inches so "5 feet, 9 inches" is 69 inches- you cannot "let it be" some other length!).

The distance, from the rock to the pivot is 9 inches. Then 69- 9= 60 inches is the distance from the pivot to the end of the crowbar where the force is being applied.
The torque due to the rock is 400(9)= 3600 "pound inches". Let F be the force applied to the end of the crowbar. Then the torque due to the end of crowbar is 65F "pound inches". In order to move the rock the force applied must be at least the "F" satisfying
60F= 3600.

(I have no idea where you got the idea that you could "Let 5 feet, 9 inches be 70.8 inches". Didn't you at least see that "5 feet 9 inches" is not a fractional amount of inches?)

Country Boy, Feb 16, 2022

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