a uniform metre rule pivoted at R, the 70cm mark. Two forces 0.1 N and 0.4 N are applied at Q, the 60cm mark and S, the 85cm mark. If the metre rule us kept in equilibrium, by the forces, calculate the weight of the metre rule.
the center of mass of the rule is at 50 cm
using the 70 cm pivot ... (20 * w) + (10 * 0.1) = 15 * 0.4
solve for w
Please explain what led to that formular..... and how did you get 20...... What I know is moment =force * distance
the rule is in equilibrium , so the moments are balanced (equal and opposite)
around the pivot point
the 20 is the distance from the pivot to the center of mass of the rule
I don't get it well
To solve this problem, we can use the principle of moments, which states that for an object to be in equilibrium, the sum of the clockwise moments must equal the sum of the counterclockwise moments.
First, let's determine the distances of the forces from the pivot point R:
- The force of 0.1 N at Q is 70 cm from R.
- The force of 0.4 N at S is 85 cm from R.
Now, let's denote the weight of the meter rule as W. Since the meter rule is balanced, the clockwise moments equal the counterclockwise moments.
The clockwise moments:
Moment of the 0.1 N force at Q = 0.1 N * 70 cm
The counterclockwise moments:
Moment of the 0.4 N force at S = 0.4 N * 85 cm
Moment of the weight W at the center of the meter rule (50 cm from R) = W * 50 cm
Since the meter rule is in equilibrium, the sum of the clockwise moments equals the sum of the counterclockwise moments:
0.1 N * 70 cm = 0.4 N * 85 cm + W * 50 cm
Now we can solve this equation to find the weight W of the meter rule:
0.1 N * 70 cm - 0.4 N * 85 cm = W * 50 cm
Multiply the force and distance values:
7 Ncm - 34 Ncm = 50 W
Combine like terms:
-27 Ncm = 50 W
Now divide both sides by 50 to solve for W:
W = -27 Ncm / 50
W ≈ -0.54 N
The weight of the meter rule is approximately 0.54 N. Note that the negative sign indicates that the weight acts in the opposite direction to the applied forces, as per convention.