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GuDViN [60]
3 years ago
11

a uniform metre rule is pivoted at its centre and weights of 5 N and 12 N are hung at the 3 cm and 5 cm marks respectively. how

far from the pivot must a 25 N weight be hung to balance the metre rule horizontally? ​
Physics
1 answer:
schepotkina [342]3 years ago
8 0

In order to balance the stick on the pivot, the total "moments" must be equal on both sides.  A "moment" is (a weight) x (its distance from the center).

for the 5N weight: Moment = (5N) x (3 cm) = 15 N-cm

for the 12N weight: Moment = (12N) x (5 cm) = 60 N-cm

Sum of the moments trying to pull the stick down on that side = 75 N-cm

Whatever we hang on the other side has to provide a moment of 75 N-cm in the other direction.  We have a 25N weight. Where should we hang it ?  

(25N) x (distance from the pivot) = 75 N-cm

Distance from the pivot = (75 N-cm) / (25 N)

<em>Distance from the pivot =  3 cm </em>

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The vertical component of the truck's velocity is: 3.13 m/s

Explanation:

You have to apply trigonometric identities for a right triangle (because the ramp can be seen as a right triangle where the speed is the hypotenuse), in order to obtain the components of the velocity vector.

The identities are:

Cosα= \frac{CA}{H}

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Where H is the hypotenuse, α is the angle, CA is the adjacent cathetus and CO is the opposite cathetus

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Let Vx represent it.

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The vertical component of the truck's velocity is:

Let Vy represent it.

In this case, CO=Vy, H=24 and α=7.5 degrees

Vy=(24)Sen(7.5)

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