Lever Problem 4 Solution

In this Algebra Lever problem solution, you again will have a problem where the distance times the weights have to equal on both sides. Also since Steve sits twice as far, he is 8 feet away since Samuel is 4 feet away and Steve sits twice as far as Samuel.

Here's the algebra equation:

65(4) + 85(8) = 200x

260 + 680 = 200x

940 = 200x

040/200 = x

94/20 = x = 4 and 14/20 or 4 and 7/10 or 4.7 feet is how far Big Billy would have to sit on the opposite side of the teeter totter across from Samuel and Steve.

Lever Problem 3 Solution

1st piece of information that's important in the Algebra lever problem is that the teeter totter was 12 feet long and the balance point was directly in the middle(which most teeter totter's are). So each side is 6 feet. This means if Bob and Susan sat on opposite ends then they were 6 feet from the fulcrum. They weigh 70 and 50 lbs respectively. On the same side as Susan 5 feet away from the center 40 lb Chrissy sat. So we got the weights and distances for everyone but Steve who weighs 35 lbs but is an unknown (x) distance from the center on the same side as Bob.

As we said before the distance times the weight of one side has to equal the other. In this case since there's more than 1 person on each side we add the respective weights and distance and set them equal to one another...

This side:

Susan = 50(6)
Chrissy = 40(5)

is equal to this side:

Bob = 70(6)
Steve = 35(x)

300 + 200 = 420 + 35x
500 = 420 + 35x

80 = 35x
x = 80/35
2 & 10/35 feet or 80/35 from the center is where Steve is located on the same side as Bob.

Just to check 500 = 420 + 35(80/35) = 420 + 80 == correct

Lever Problem 2 Solution

In lever problems you are trying to get the weight * the distance of one side to equal the weight times the distance of the other. The fulcrum would be considered the balance point, and since a weight of 100 lbs is on that side we can see that the weight of 100 * 2 = 200 has to equal the distance of 8 (10 - 2) times a certain amount of weight which were trying to figure out for the other side. So algebraically we have:

100 * 2 = 8 * x

200 = 8x

x = 200/8

x = 25

This means that a weight of 25 lbs would have to be applied to one side of the lever to lift the 100 lbs. As far as levers are concerned this makes sense. You definitely want to exert less force on one end that the weight of object your trying to move!

Lever Problem 1 Solution

With any type of lever problem it's important to realize that the weight of object 1 TIMES the distance of that object from the center(or fulcrum) is going EQUAL the weight of object 2 TIMES the distance of that object from the center(fulcrum).

So in this problem you know the to weights are Sam at 100 lbs and James at 150 lbs, and the distance from the fulcrum is known for Sam which is 8(feet). However, you don't know James' distance so thats what were trying to find.

So set them equal

100 * 8 = 150 * x

800 = 150x

x = 800/150

x = 5 1/3 (feet) or 5 feet 4 inches (since 1(ft)/3 * 12 inches/ft = 4 inches)