a uniform meter stick supported at a 25-cm mark balances when a 1 kg rock is suspended at the 0 cm end. What is the mass of the meter stick??
2006-12-03 23:25:06 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Balance the forces around the lever point at .25 m. so 1 kg * g * 25 equals the force of gravity at the center of mass of the meter stick or 0.5 metes times the distance to the lever point. Therefore the meter stick also weighs 1 kilogram
2006-12-03 23:41:08 · answer #1 · answered by Anonymous · 1⤊ 0⤋
Let's look at the meter stick in halves. The first half (0cm - 50cm) is "balanced", since the stick is being supported at the 25-cm mark. So we can ignore that for now.
The other half of the stick (from the 50-cm mark to the 100-cm mark) is the main thing to look at. It is extended out on the "long side" of the balance, and its mass is being counterweighted by the 1kg rock suspended at the 0-cm end on the short side.
The center of mass of the rock is 25 cm from the fulcrum. The center of mass of the extended half of the uniform stick would be at the midpoint of that segment of the stick, which would be the 75-cm mark, which is 50 cm from the fulcrum. Therefore, the mass of that half of the stick would be (25 * 1kg) / 50 = 1/2 kg. Since half of the stick is 1/2 kg, the entire stick must be 1kg -- the same mass as the rock, in this instance.
2006-12-03 23:46:33 · answer #2 · answered by PM 3 · 0⤊ 0⤋
nckobra40 gave you the first right answer. But I didn't really like anyone's explanation.
The center of mass of the meter stick is at the 50 cm point, 25 cm to the right of the pivot.
The system is balanced, so
CCW torque = CW torque
1 kg*g* 25 cm = m*g*25 cm
m = 1 kg
2006-12-04 05:03:53 · answer #3 · answered by sojsail 7 · 1⤊ 1⤋
The 25-cm mark will be where the pivot is.
The weight of the rock= 1*10 = 10 N
The weight of the meter stick is in the middle of the stick which is at 50 cm mark. (which is 25 cm from the pivot)
If you draw a picture out, you will see that the rock creates an anti-clockwise moment while the weight of the meter stick creates a clockwise moment.
Taking moments about the pivot, by Principle of Moments,
clockwise moments= anti-clockwise moments
therefore,
weight of stick*distance of the weight of stick from pivot=weight of rock*distance of the weight of rock from pivot
weight of stick*25cm=10N*25cm
weight of stick=10N*25cm / 25cm
weight of stick= 10N
Thus,
mass of stick= 10N/10
= 1 kg
2006-12-04 01:35:36 · answer #4 · answered by mini_gal 3 · 0⤊ 0⤋
You know that 1Kg + .25 of the stick's mass balances .75 of the stick's mass (m). Write that as an equation
1 +.25m = .75m
1=.75m-.25m
The stick weighs 2 Kg
2006-12-03 23:40:00 · answer #5 · answered by Iridflare 7 · 0⤊ 0⤋
since it is a uniform stick teh mass is centered at 50cm mark. But ti only has .75 of its mass as 25cm is on the other side of the fulcrum
1kg(25cm)=.75(Xkg)(25cm)
Eliminate the 25cm it is on both sides
1kg=.75(xkg)
1/.75=xKg
1.33kg=mass of meter stick
2006-12-03 23:46:54 · answer #6 · answered by shadouse 6 · 0⤊ 0⤋
Seems simple enough:
We are saying that the mass of one-quarter of the stick plus 1kg must equal the mass of the other three-quarters of the stick, so:
1kg + 0.25X = 0.75X
1kg = (0.75 - 0.25)X
1kg = 0.5X
X = 2kg
(where X is the mass of the stick).
2006-12-04 01:47:59 · answer #7 · answered by Mawkish 4 · 0⤊ 1⤋