a uniform rod 1m long with weight 6N can be supported in a horizontal position on a sharp edge with weights of 10N and 15N suspended from its ends. What is the position of point of balance?
Answers
Answer:
x = 18/31m from 10N = 0.5806m
Explanation:
Imagine a rod with two endpoints A and B. The 10N Weight is on A end and 15N weight is on B end.
now let a point C at a distance x from A where the rod is balanced.
taking moment about point x we get
10*x = (0.5-x)*6 + (1-x)*15
solve this equation
x = 18/31m from A = 0.5806m
Answer: The answer is 0.41 m
Explanation: This question really is about torque. Let us call the position of balance as B. Let L is the length of the rod = 100 cm; let the 10 N force Force 1 hanged at the left side at 0 cm; let Force 2 hanged at the right side of P at 100 cm. Let X is the distance to the left of point B and let 100 - X is the distance to the right of point B.
You can find the answer Force 1 * X = Force 2 * (100 - X)
10 N * X cm = 15 N * (100 cm - X cm)
10 X N * cm = 1500 N * cm - 15 X N * cm
10 X N * cm + 15 X N * cm = 1500 N * cm
25 X N * cm = 1500 N * cm
25 X = 1500
X = 1500 / 25
X = 60 cm
The position of the point of balance is at the 6 m from The 10 N force Force 1 or at 4 m from the 15 N force Force 2.