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QuestionA 6 m6 \mathrm{~m} beam pivots at the center. A 5 N5 \mathrm{~N} weight is 2 m2 \mathrm{~m} from the pivot. What weight balances it?

Studdy Solution

STEP 1

Assumptions1. The beam is6m in length. The beam pivots at the point of the support3. A5 weight is placed on one end, which ism from the pivot point4. We need to find the weight that should be placed on the other end to balance the beam5. The beam will balance when the moments on both sides of the pivot are equal. The moment is calculated as the product of the force and the distance from the pivot point.

STEP 2

First, we need to calculate the moment caused by the5 weight. The moment is calculated as the product of the force and the distance from the pivot point.
Moment=ForcetimesDistanceMoment = Force \\times Distance

STEP 3

Now, plug in the given values for the force and distance to calculate the moment.
Moment=5times2mMoment =5 \\times2m

STEP 4

Calculate the moment caused by the weight.
Moment=times2m=10NmMoment = \\times2m =10Nm

STEP 5

To balance the beam, the moment on the other side of the pivot point must be equal to10Nm. Let's denote the weight that should be placed on the other end as WW and the distance from the pivot point as $$. We know that $ =m -2m =4m$.
10Nm=Wtimes4m10Nm = W \\times4m

STEP 6

Now, we can solve for WW by dividing both sides of the equation by4m.
W=10Nm4mW = \frac{10Nm}{4m}

STEP 7

Calculate the weight that should be placed on the other end to balance the beam.
W=10Nm4m=2.5W = \frac{10Nm}{4m} =2.5The weight that should be placed on the other end to balance the beam is2.5.

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