A uniform board 12 ft long weighs 10 lbs. Two feet from the left is centered a 3 lbs. structure. How far from the left end is the balance point i.e. the point at which if you applied 13 lbs. upward force would balance the board and structure??
total weight = 10 + 3 = 13
13 x = 6(10) + 3(2) = 66
x = 66/13 = 5.08 ft
thanks so much
You are welcome :)
To find the balance point of the board and structure, we need to calculate the distance from the left end.
First, let's calculate the total weight of the board and structure. The board weighs 10 lbs and the structure weighs 3 lbs, so the total weight is 10 lbs + 3 lbs = 13 lbs.
Now, let's calculate the torque on each side of the balance point. Torque is the force applied multiplied by the distance from the pivot point (in this case, the left end of the board).
On the left side of the balance point:
Torque = weight × distance
Torque = 10 lbs × x ft, where x is the distance from the left end
On the right side of the balance point:
Torque = weight × distance
Torque = 3 lbs × (12 - x) ft, where (12 - x) is the distance from the right end (12 ft) since the structure is centered 2 ft from the left end.
For the board and structure to be balanced, the torque on the left and right sides should be equal.
Therefore, we can set up the equation:
10 lbs × x ft = 3 lbs × (12 - x) ft
Now, let's solve for x:
10x = 3(12 - x)
10x = 36 - 3x
13x = 36
x = 36 / 13
x ≈ 2.77 ft
So, the balance point is approximately 2.77 ft from the left end of the board.