Question

A metre stick has its centre of mass at the 50cm mark, and weighs 0.92n. A 2.00n weight is stuck to the 10cm mark with massless glue. About which point will the ruler balance?

Answer 1

The meter stick will balance at approximately 22.60cm from the 0cm end. This point is found by setting the torque produced by the meter stick's weight and the additional weight to be equal and opposite about the balance point, and solving for the position of the balance point.

Step-by-step explanation:

To find the point about which the meter stick will balance with the added weight, we need to consider the concept of torque balance. Torque is the product of the force applied and the distance from the pivot point. To achieve balance, the sum of the torques around any point must be zero. In this case, the meter stick has its own weight (0.92N) acting at the 50cm mark, and there is a 2.00N weight at the 10cm mark.

We set up the equation for balance about the point x (where x is the distance from the 0cm end):

• Left side torque: (2.00N) × (x - 10cm)
• Right side torque: (0.92N) × (50cm - x)

Set the left side torque equal to the right side torque and solve for x:

(2.00N) × (x - 10cm) = (0.92N) × (50cm - x)

Expanding and simplifying, we get:

2.00N × x - 20.00N/cm = 46.00N/cm - 0.92N × x

2.92N × x = 66.00N/cm

x = 66.00N/cm / 2.92N

x ≈ 22.60cm

Therefore, the meter stick will balance at approximately 22.60cm from the 0cm end.

Answer 2

To determine the new balance point of the ruler with an additional weight attached at the 10cm mark, we use the principle of moments. The balance point is calculated by setting the clockwise moments equal to the anti-clockwise moments, which results in the ruler balancing at approximately the 22.6cm mark.

Step-by-step explanation:

To find the point at which the ruler will balance, we can use the principle of moments, which states that for an object to be in equilibrium, the clockwise moments must equal the anti-clockwise moments. The given problem involves a metre stick that has its centre of mass at the 50cm mark, and it weighs 0.92n. A 2.00n weight is stuck to the 10cm mark with massless glue.

To balance the ruler, the moments around the pivot point must be equal. This can be expressed as:

(Weight1 × Distance1) = (Weight2 × Distance2)

where Weight1 is the weight of the metre stick and Distance1 is the distance from the pivot point to the 50cm mark, and Weight2 is the weight of the attached object and Distance2 is the distance from the pivot point to the 10cm mark.

In this case, the weight of the metre stick (0.92n) is balanced at the 50cm mark, so we can consider the 50cm mark as the pivot point:

(0.92n × 0cm) = (2.00n × Distance)

Notice that since the metre stick is already balanced at its centre of mass, the distance for the 0.92n force is 0cm, effectively making this side of the equation zero. The entire moment will therefore be on the side of the weight:

(2.00n × Distance) = 0

However, since the metre stick is already balanced at the 50cm mark without the additional weight, the addition of a weight at the 10cm mark will make the stick no longer balance at the 50cm mark. To find the new balance point (x), we can set up a balance of moments equation where Distance1 is now (x - 10cm) and Distance2 is (50cm - x) since the origin is taken at the 10cm mark where the new weight is added:

(0.92n × (50cm - x)) = (2.00n × (x - 10cm))

Solving for x gives:

0.92n × (50 - x) = 2.00n × (x - 10)

46 - 0.92x = 2x - 20

Adding 0.92x to both sides and adding 20 to both sides gives:

66 = 2.92x

Dividing both sides by 2.92 gives:

x = 22.6cm (approximately)

Therefore, the ruler will balance at approximately the 22.6cm mark.