Force — ICSE Class 10 Physics Important Questions

Answer: (b) $\text{N m}$N m.

Moment $=$= force $\times$× perpendicular distance $=\text{N}\times\text{m}=\text{N m}$=N×m=N m. Although $\text{N m}$N m equals a joule dimensionally, the moment is not energy, so it is written $\text{N m}$N m, not $\text{J}$J.

Answer: (a) decreases the force needed by increasing the moment arm.

Moment $=F\times d$=F× d. Pushing far from the hinge gives a larger perpendicular distance $d$d, so a smaller force $F$F produces the same turning moment.

Answer: (b) only rotation.

The two equal and opposite forces of a couple give zero resultant force (no translation) but a net moment, so the body only rotates.

Answer: (b) the $70\text{ cm}$70 cm mark.

Anticlockwise moment $=20\text{ gf}\times(50-20)=20\times30=600\text{ gf cm}$=20 gf×(50-20)=20×30=600 gf cm. For balance, $30\times d=600\Rightarrow d=20\text{ cm}$30× d=600 d=20 cm from the pivot, i.e. at the $50+20=70\text{ cm}$50+20=70 cm mark.

Answer: (c) The centre of gravity of a ring is indeed at its centre, which lies in the empty space, so A is true. R is false because the centre of gravity need not lie within the material (a ring is the standard example).

The moment of a force about a point is the turning effect of the force about that point, measured by the product of the force and the perpendicular distance of its line of action from the point.

$\text{Moment}=F\times d$Moment=F× d

It depends on: (i) the magnitude of the force $F$F, and (ii) the perpendicular distance $d$d of the line of action of the force from the pivot.

The perpendicular distance between the two forces of the couple equals the diameter, $d=2\text{ m}$d=2 m.

Moment of couple $=$= one force $\times$× perpendicular distance
$\tau=F\times d=5\text{ N}\times2\text{ m}=10\text{ N m}.$=F× d=5 N×2 m=10 N m.

The moment of the couple is $10\text{ N m}$10 N m.

Since the rule is uniform and pivoted at its centre ($50\text{ cm}$50 cm mark), its weight acts at the centre through the pivot, so its moment is zero and it can be ignored.

Anticlockwise moment of $150\text{ gf}$150 gf about the pivot:
$150\text{ gf}\times(50-10)\text{ cm}=150\times40=6000\text{ gf cm}.$150 gf×(50-10) cm=150×40=6000 gf cm.

Let the $50\text{ gf}$50 gf weight hang at a distance $d$d on the other side. By the principle of moments:
$50\times d=6000\Rightarrow d=120\text{ cm}.$50× d=6000 d=120 cm.

Since $d=120\text{ cm}$d=120 cm exceeds the rule's length, balancing with only $50\text{ gf}$50 gf is not possible; a larger weight or an additional support is required. (This shows the importance of checking the physical range of the answer.)

Principle of moments: When a body is in rotational equilibrium, the algebraic sum of the moments of all forces about any point is zero; i.e. the sum of the anticlockwise moments equals the sum of the clockwise moments about the pivot:
$\sum\tau_{acw}=\sum\tau_{cw}.$_acw=_cw.

Physical balance: The beam is pivoted at its centre with two identical pans hung at equal distances $d$d on either side of the fulcrum. When the body of weight $W$W is on one pan and standard weights $W'$W' on the other, balance is reached when their moments are equal:
$W\times d=W'\times d\Rightarrow W=W'.$W× d=W'× d W=W'.

Since the arm lengths are equal, the unknown weight equals the total standard weights, giving the body's weight directly.

The rule is uniform, so its whole weight $W$W acts at its centre, the $25\text{ cm}$25 cm mark.

Pivot (knife-edge) is at the $29\text{ cm}$29 cm mark.

Taking moments about the pivot:

Anticlockwise moment (due to $20\text{ gf}$20 gf at $10\text{ cm}$10 cm):
$20\times(29-10)=20\times19=380\text{ gf cm}.$20×(29-10)=20×19=380 gf cm.

Clockwise moment (due to weight $W$W at $25\text{ cm}$25 cm... note $25<29$25<29, so this is on the same side): Re-checking, both the $10\text{ cm}$10 cm load and the $25\text{ cm}$25 cm centre lie to the left of the pivot, giving anticlockwise moments, which cannot balance. Hence the weight of the rule acts at $25\text{ cm}$25 cm, to the left, and the balancing requires the load and the rule's weight on opposite sides. The load at $10\text{ cm}$10 cm is left of the pivot; the centre of gravity at $25\text{ cm}$25 cm is also left. For equilibrium the knife-edge must lie between them, so the pivot at $29\text{ cm}$29 cm gives:
$W\times(29-25)=20\times(29-10)$W×(29-25)=20×(29-10)
$W\times4=380\Rightarrow W=95\text{ gf}.$W×4=380 W=95 gf.

The weight of the rule is $95\text{ gf}$95 gf.

(a) A single force can produce translation, rotation, or both, and has a resultant that can move the body. A couple is a pair of equal, opposite, parallel forces whose lines of action differ; its resultant force is zero, so it produces pure rotation only.

(b) A body is in equilibrium when it is either at rest or in uniform motion, with no change in its state. The two conditions are:

1. The resultant of all forces acting on it is zero ($\sum F=0$F=0).
2. The resultant moment of all forces about any point is zero ($\sum\tau=0$=0).

(c) The pivot is at the centre, so each end is $2\text{ m}$2 m from the pivot. The boy sits at one end, $2\text{ m}$2 m away.

Moment of boy $=400\text{ N}\times2\text{ m}=800\text{ N m}$=400 N×2 m=800 N m (say anticlockwise).

Let the girl sit at distance $d$d on the other side. By the principle of moments:
$500\times d=800\Rightarrow d=1.6\text{ m}.$500× d=800 d=1.6 m.

The girl must sit $1.6\text{ m}$1.6 m from the pivot (i.e. $0.4\text{ m}$0.4 m from her end).

The forces are: weight of rule $60\text{ gf}$60 gf acting downward at $50\text{ cm}$50 cm, load $40\text{ gf}$40 gf downward at $30\text{ cm}$30 cm, and upward tensions $T_A$T_A at $0\text{ cm}$0 cm and $T_B$T_B at $100\text{ cm}$100 cm.

Taking moments about A (at $0\text{ cm}$0 cm): clockwise moments $=$= anticlockwise moment of $T_B$T_B.
$T_B\times100=60\times50+40\times30$T_B×100=60×50+40×30
$T_B\times100=3000+1200=4200\Rightarrow T_B=42\text{ gf}.$T_B×100=3000+1200=4200 T_B=42 gf.

Total upward force $=$= total downward force:
$T_A+T_B=60+40=100\text{ gf}$T_A+T_B=60+40=100 gf
$T_A=100-42=58\text{ gf}.$T_A=100-42=58 gf.

Tension in string A $=58\text{ gf}$=58 gf and in string B $=42\text{ gf}$=42 gf.

(i) The principle of moments: in rotational equilibrium, the sum of anticlockwise moments about the pivot equals the sum of clockwise moments.

(ii) Anticlockwise moment $=100\text{ gf}\times(50-20)\text{ cm}=100\times30=3000\text{ gf cm}.$=100 gf×(50-20) cm=100×30=3000 gf cm.

(iii) The clockwise moment must equal $3000\text{ gf cm}$3000 gf cm. The unknown acts at $80\text{ cm}$80 cm, i.e. $30\text{ cm}$30 cm from the pivot:
$W\times30=3000\Rightarrow W=100\text{ gf}.$W×30=3000 W=100 gf.

(iv) Balancing the rule at its centre first ensures that the rule's own weight acts through the pivot, giving zero moment, so it does not affect the readings and the two hung weights alone are compared.