Question

(1) A uniform metre rule balances horizontally on a knife edge placed at the 58cm mark when a weight of 20gf is suspended from one end.
(a) Draw a diagram of the arrangement.
(b) What is the weight of the rule?
(2) A uniform metre scale can be balanced at the 70cm mark when a mass 0.05kg is hung from the 94cm mark. Find the mass of the metre scale.
(3) A truck weighing 1000 kgf changes its speed from 36 km/hr to 72km/hr in 2 minutes. Calculate (a) the work done by the engine and (b) its power. (g= 10metre per second square.

Answer 1

Answer for a)
the diagram would be something like this :
make a meter scale and on the last end place the weight of 20 gf. At 58 cm mark
the fulcrum and at 50 cm mark the weight of scale (downward direction both can be depicted by arrows.
Answer for b)
sum of anticlockwise force= sum of clockwise force
hence let the weight of meter scale be x
x * distance from fulcrum( here 8 cm) = 20 * 42
x=105 gf

Answer 2

1-(b) from the principle of moment
.... Anticlockwise moment= clockwise moment
W*(58-50)=20*(100-58)
W*8=20*42
W=
$\frac{20 \times 42}{8}$
....... = 105 gf