A uniform metal bar is 5.00 m long and has mass 0.300 kg. The bar is pivoted on a narrow support that is 2.00 m from the left-ha
nd end of the bar. What distance x from the left-hand end of the bar should an object with mass 0.900 kg be suspended so the bar is balanced in a horizontal position?
1 answer:
Explanation and answer:
This type of question can be clarified and sometimes solved by drawing a proper diagram or sketch. (see below)
Solution:
Since we do not know the reaction of the support, we can take moments about the support (thereby eliminating its involvement).
CCW moment = 0.900(5.00/2 - x) kg-m
CW moment = 0.300*(5.00/2-2.00)) = 0.150 kg-m
At equilibrium, CCW moment = CW moment, so
0.900(2.50-x) = 0.150
Expand and solve
2.25 - 0.900x = 0.150
0.900x = 2.25-0.15 = 2.10
x = 2.10 / 0.900 = 2.33 m (to nearest cm)
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Answer:
103.5 meters
Explanation:
Given that a stunt person has to jump from a bridge and land on a boat in the water 22.5 m below. The boat is cruising at a constant velocity of 48.3 m/s towards the bridge. The stunt person will jump up at 6.45 m/s as they leave the bridge.
The time the person will jump to a certain spot under the bridge can be calculated by using the formula below:
h = Ut + 1/2gt^2
since the person will fall under gravity, g = 9.8 m/s^2
Also, let assume that the person jump from rest, then, U = 0
Substitute h, U and g into the formula above
22.5 = 1/2 * 9.8 * t^2
22.5 = 4.9t^2
22.5 = 4.9t^2
t^2 = 22.5/4.9
t^2 = 4.59
t =
t = 2.143 seconds
From definition of speed,
speed = distance /time
Given that the boat is cruising at a constant velocity of 48.3 m/s towards the bridge, substitute the speed and the time to get the distance.
48.3 = distance / 2.143
distance = 48.3 * 2.143
distance = 103.5 m
Therefore, the boat should be 103.5m away from the bridge at the moment the stunt person jumps?