The meterstick shown to the right is 100 cm long. It is free to pivot around its center of gravity (CG), which is at the 50-cm mark. There is a 25.0-N block hanging from the 80-cm mark. Decide where each of the other blocks should be placed, one at a time, to balance out the 25.0-N block. At what mark on the meter stick would you place a 16.0-N block to balance the 25.0-N block?

The 16.0-N block should be placed at the 64-cm mark on the meter stick in order to balance the 25.0-N block.

To balance the 25.0-N block, we need to ensure that the torques on both sides of the meterstick are equal. The torque is calculated by multiplying the force applied by the distance from the pivot point.

1. Firstly, let's calculate the torque exerted by the 25.0-N block at the 80-cm mark:
Torque = Force x Distance
Torque = 25.0 N x 80 cm = 2000 N.cm

2. Since the meterstick is balanced around its center of gravity at the 50-cm mark, the torque exerted by the 16.0-N block must be equal but opposite to balance the system.

3. Let's denote the position of the 16.0-N block as "x" cm from the 50-cm mark. The torque exerted by the 16.0-N block is given by:
Torque = Force x Distance
Torque = 16.0 N x (50 cm - x)

4. Set the torques of both blocks equal to each other and solve for "x" to find the position where the 16.0-N block should be placed:
2000 N.cm = 16.0 N x (50 cm - x)

5. Expanding the equation, we get:
2000 = 16x(50 - x)

6. Simplifying, we have:
2000 = 800 - 16x^2

7. Rearranging the equation, we get:
16x^2 = 800 - 2000
16x^2 = -1200

8. Dividing by 16:
x^2 = -75

9. Since we cannot have a negative position on the meterstick, it is not possible to place the 16.0-N block in a position that will balance the 25.0-N block.

To balance out the 25.0-N block, the torque on one side of the meterstick should equal the torque on the other side. The torque is calculated by multiplying the force (in newtons) by the distance (in meters) of the force from the pivot point.

In this case, the torque on one side is the product of the force on that side and the distance of that force from the pivot point. The torque on the other side is the product of the force on that side (16.0 N) and the distance of that force from the pivot point (x).

Since the meterstick is balanced, we can equate the torque on both sides:

Torque on side with 25.0-N block = Torque on side with 16.0-N block

(25.0 N) * (80 cm) = (16.0 N) * (x)

To find x, we need to solve this equation.

First, let's convert the distance from centimeters to meters:

80 cm = 80/100 m = 0.8 m

Rearranging the equation:

(25.0 N) * (0.8 m) = (16.0 N) * (x)

Now, we can solve for x:

(25.0 N * 0.8 m) / 16.0 N = x

x = 1.25 m

So, to balance the 25.0-N block, the 16.0-N block should be placed at the 1.25-meter mark on the meterstick.