A boy of mass 23 kg sits on a seesaw 1.5 m from the axis of rotation. At what
distance from the axis of rotation must his friend, whose mass is 21 kg, be
positioned to balance the seesaw? [6 marks]
Equate moments about fulcrum:
23*1.5=21*x
Solve for x.
I think its 1.64 metres
To balance the seesaw, the torque on one side must be equal to the torque on the other side. The torque is given by the formula:
Torque = force × distance
Since the force is the weight of the person, the torque can be written as:
Torque = weight × distance
The weight is given by the formula:
Weight = mass × gravity
where gravity is approximately equal to 9.8 m/s².
Let's calculate the torque for both sides.
For the boy:
Weight of the boy = mass of the boy × gravity
Weight of the boy = 23 kg × 9.8 m/s²
Weight of the boy ≈ 225.4 N
Torque for the boy = weight of the boy × distance
Torque for the boy = 225.4 N × 1.5 m
Torque for the boy ≈ 338.1 N·m
For the friend:
Weight of the friend = mass of the friend × gravity
Weight of the friend = 21 kg × 9.8 m/s²
Weight of the friend ≈ 205.8 N
Let x be the distance from the axis of rotation where the friend is positioned.
Torque for the friend = weight of the friend × distance
Torque for the friend = 205.8 N × x
Since the seesaw is balanced, the torque for the boy and the torque for the friend must be equal:
Torque for the boy = Torque for the friend
338.1 N·m = 205.8 N × x
Now, let's solve for x:
x = 338.1 N·m / 205.8 N
x ≈ 1.64 m
Therefore, the friend must be positioned approximately 1.64 meters from the axis of rotation to balance the seesaw.
To find the distance from the axis of rotation where the friend should be positioned to balance the seesaw, we need to consider the principle of torque, which states that for an object to be balanced around a pivot point, the sum of the clockwise torques must be equal to the sum of the counterclockwise torques.
In this case, the torque is calculated as the product of the force applied and the distance from the pivot point. Since the force due to gravity is acting downward, the clockwise torque is considered to be negative while the counterclockwise torque is positive.
Let's assume the distance where the friend should be positioned is 'x' meters from the axis of rotation.
For the boy, the clockwise torque would be:
Clockwise Torque (Boy) = (-23 kg) * (9.8 m/s^2) * (1.5 m) = -338.7 Nm
For the friend, the counterclockwise torque would be:
Counterclockwise Torque (Friend) = (21 kg) * (9.8 m/s^2) * (x m)
Since the seesaw needs to be balanced, we can equate the torques:
-338.7 Nm = (21 kg) * (9.8 m/s^2) * (x m)
Solving for x, we have:
x = (-338.7 Nm) / (21 kg * 9.8 m/s^2)
x = -1.7 m
The negative sign indicates that the friend should be positioned 1.7 meters to the left of the axis of rotation to balance the seesaw.