The diagram below shows a uniform bar supported at the middle point O. A weight of $40$ gf is placed at a distance $40$ cm to the left of point O. How can you balance the bar with a weight of $80$ gf ?

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Solution

It is known that
$τ = \to F \times \to r$
moment of force = force (or weight) multiplied by the arm length from the pivot, in this case , that is the middle point "O".

Place $80 g f$ to the right of O on the bar at a distance "X". Then
$80 g f \times x = 40 g f \times 40 c m$
$x = 20 c m$

so $x = 20 c m$ .
we can balance the bar with a weigh by placing the 80gf at 20 cm to the right of O.

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The diagram below shows a uniform bar supported at the middle point O. A weight of 40 gf is placed at a distance of 40 cm to the left of the point O. How can you balance the bar with a weight of 80 gf?

Fig. 1.33 shows a uniform metre rule placed on a fulcrum at its mid-point O and having a weight 40 gf at the 10 cm mark and a weight of 20 gf at the 90 cm mark. (i) Is the metre rule in equilibrium ? If not, how will the rule turn ? (ii) How can the rule be brought in equilibrium by using an additional weight of 40 gf ?

The diagram in the figure shows a uniform metre rule weighing 100 gf, pivoted at its centre O. Two weights 150 gf and 250 gf hang from the points A and B of the metre rule such that OA = 40 cm and OB = 20 cm. Calculate:

(i) the total anticlockwise moment about O,

(ii) the total clockwise moment about O,

(iii) the difference of anticlockwise and clockwise moments,

and (iv) the distance from O where a 100 gf weight should be placed to balance the metre rule.