Question 1

Question 1A, B and C are the three forces each of magnitude 4N acting in the plane of paper as shown in figure. The point O lies in the same plane.(i) Which force has the least moment about O? Give reason.(ii) Which force has the greatest moment about O? Give reason.(iii) Name the forces producing (a) clockwise, (b) anticlockwise moments.(iv) What is the resultant torque about the point O?

Accepted Answer

(i)We know that,Moment of force = Force × Perpendicular distanceWe can see from the given figure that vector C has least perpendicular distance from point O.Hence, vector C will have least moment about O.(ii) We know that,Moment of force = Force × Perpendicular distanceWe can see from the given figure that vector A has greatest perpendicular distance from point OHence, vector A will have greatest moment about O.(iii)(a) Clockwise moments are produced by vectors A and B.Reason : If turning effect on the body is clockwise then moment of force is called clockwise moment and it is taken as a negative value.(b) Anticlockwise moment is produced by vector C.Reason : If turning effect on the body is anticlockwise then moment of force is called anticlockwise moment and it is taken as a positive value.(iv) Sum of torques due to vectors A, B and C = Resultant torque about point O−(4×0.9)−(4×0.8)+(4×0.6)Nm=−3.6−3.2+2.4Nm=−4.4Nm-(4 \times 0.9) - (4 \times 0.8) + (4 \times 0.6)Nm \[0.5em] = -3.6 - 3.2 + 2.4 Nm \[0.5em] = -4.4 Nm−(4×0.9)−(4×0.8)+(4×0.6)Nm=−3.6−3.2+2.4Nm=−4.4NmSo, the resultant torque is 4.4Nm (clockwise direction).

Question 2

Question 4A mechanic applies a force of 100 N at the end of a spanner to tighten a bolt. The length of the spanner is 0.5 metres. How much torque is exerted on the bolt? If the mechanic increases the force applied by 20% but holds the spanner closer at 0.4 metres, would it result in an increase or decrease of torque?

Accepted Answer

In first case,F1 = 100 Nr1 = 0.5 mTorque 1 = r1 x F1 = 0.5 x 100 = 50 Nm∴ Torque exerted on the bolt = 50 NmIn second case,F2 = F1 + 20% x F1 = F1 + 0.2 x F1 = 1.2 F1 = 1.2 x 100 = 120 Nr2 = 0.4 mTorque 2 = r2 x F2 = 0.4 x 120 = 48 NmHere, Torque 1 > Torque 2Torque 1 - Torque 2 = 50 Nm - 48 Nm = 2 Nm∴ The torque will decrease by 2 Nm

Question 3

Question 5A wheel of diameter 2 m is shown in figure with axle at O. A force F = 2 N is applied at B in the direction shown in figure.Calculate the moment of force about(i) the centre O, and(ii) the point A.

Accepted Answer

We know from the question,F = 2 NDiameter = 2mSo, Radius or OB = 1m(i) Moment of force at O = F x rSubstituting the values of F and r we get,2×1=2Nm2 	imes 1 = 2Nm2×1=2Nm (ii) Moment of force at A = F x rSubstituting the values of F and r we get,2×2=4Nm2 	imes 2 = 4Nm2×2=4Nm

Question 4

Question 6The diagram shows two forces F1 = 5N and F2 = 3N acting at points A and B of a rod pivoted at a point O, such that OA = 2m and OB = 4mCalculate:(i) the moment of force F1 about O.(ii) the moment of force F2 about O.(iii) total moment of the two forces about O.

Accepted Answer

Given,OA = 2mOB = 4mF1 = 5NF2 = 3NAs we know,Moment of force = F x rSubstituting the values of F and r(i) Moment of force F1 about O5×2=10Nm5 	imes 2 = 10Nm5×2=10NmTherefore, moment of force F1 about O is 10Nm (anticlockwise)(ii) Moment of force F2 about O3×4=12Nm3 	imes 4 = 12Nm3×4=12NmTherefore, moment of force F2 about O is 12Nm (clockwise)(iii) Total moment of two forces about midpoint is12−10=2Nm12 - 10 = 2Nm12−10=2NmTherefore, total moment of two forces about O is 2Nm (clockwise)

Question 5

Question 7Two forces each of magnitude 10N act vertically upwards and downwards respectively at the two ends A and B of a uniform rod of length 4m which is pivoted at its mid-point O as shown. Determine the magnitude of resultant moment of forces about the pivot O.

Accepted Answer

Given,Ab = 4mOA = 2mOB = 2mForce at A = 10NForce at B = 10NAs we know ,Moment of force = F x rSubstituting the values of F and rMoment of force about o at point A10×2=20Nm clockwise10 \times 2 = 20Nm \text{ clockwise}10×2=20Nm clockwiseMoment of force about o at point B10×2=20Nm clockwise10 \times 2 = 20Nm \text{ clockwise}10×2=20Nm clockwiseTotal moment of forces about the centre O20+20=40Nm20 + 20 = 40Nm \[0.5em]20+20=40NmTotal moment of force about the pivot O is 40 Nm (clockwise)

Question 6

Question 8Figure shows two forces each of magnitude 10N acting at the points A and B at a separation of 50 cm, in opposite directions. Calculate the resultant moment of the two forces about the point (i) A, (ii) B and (iii) O, situated exactly at the middle of the two forces.

Accepted Answer

Given,Force at A = 10NForce at B = 10NDistance between A and B = 50cm = 0.5m(i) Resultant moment of two forces at point A is10×0.5=5Nm clockwise10 \times 0.5 = 5Nm \text{ clockwise} \[0.5em]10×0.5=5Nm clockwiseTherefore, resultant moment of two forces at point A is 5 Nm (clockwise)(ii) Resultant moment of two forces at point B is10×0.5=5Nm clockwise10 \times 0.5 = 5Nm \text{ clockwise} \[0.5em]10×0.5=5Nm clockwiseTherefore, resultant moment of two forces at point B is 5 Nm (clockwise)(iii) Perpendicular distance of point O from either of the forces F = 10N is 0.25 mMoment of force F at point A about O10×0.25=2.5Nm clockwise10 \times 0.25 =2.5Nm \text{ clockwise} \[0.5em]10×0.25=2.5Nm clockwiseMoment of force F at point B about O10×0.25=2.5Nm clockwise10 \times 0.25 =2.5Nm \text{ clockwise} \[0.5em]10×0.25=2.5Nm clockwiseResultant moment of two forces about o is2.5+2.5=5Nm clockwise2.5 + 2.5 = 5Nm \text{ clockwise} \[0.5em]2.5+2.5=5Nm clockwiseTherefore, Resultant moment of two forces about o is 5Nm (clockwise)

Question 7

Question 9A steering wheel of diameter 0.5m is rotated anti-clockwise by applying two forces each of magnitude 6N. Draw a diagram to show the application of forces and calculate the moment of forces applied.

Accepted Answer

Moment of couple = either force × couple arm= 6 x 0.5 = 3 Nm

Question 8

Question 10A uniform metre rule is pivoted at its mid-point. A weight of 50 gf is suspended at one end of it. Where should a weight of 100gf be suspended to keep the rule horizontal?

Accepted Answer

Let us assume that a 50 gf weight produces an anticlockwise moment about the middle point ( 50 cm ).Now, if a weight of 100 gf produces a clockwise moment about the middle point and d cm be the distance from the middle.As we know, the principle of moments states thatAnticlockwise moment = Clockwise moment.50gf×50cm=100gf×dcm⇒d=50×50100⇒d=25cm50 gf \times 50cm = 100gf \times d cm \[0.5em] \Rightarrow d = \dfrac{50\times50}{100}\[0.5em] \Rightarrow d = 25 cm50gf×50cm=100gf×dcm⇒d=10050×50​⇒d=25cmTherefore, a weight of 100gf will be suspended at a distance of 25cm to keep the ruler balanced.

Question 9

Question 11A uniform metre rule balances horizontally on a knife edge placed at the 58 cm mark when a weight of 20gf is suspended from one end. (i) Draw a diagram of the arrangement. (ii) What is the weight of the rule?

Accepted Answer

(i) Diagram of the arrangement is shown below:(ii) As we know, the principle of moments states thatAnticlockwise moment = Clockwise moment.W×(58−50)=20gf×(100−58)⇒W=20×428⇒W=105gfW \times (58-50) = 20gf \times (100-58) \[0.5em] \Rightarrow W = \dfrac{20\times42}{8} \[0.5em] \Rightarrow W = 105 gfW×(58−50)=20gf×(100−58)⇒W=820×42​⇒W=105gf

Question 10

Question 12The diagram shows a uniform bar supported at the middle point O. A weight of 40 gf is placed at a distance 40cm to the left of the point O. How can you balance the bar with a weight of 80 gf?

Accepted Answer

From the given figure we can see that,Anticlockwise moment = 40 gf × 40 cmClockwise moment = 80 gf × d cmAs we know, the principle of moments states thatAnticlockwise moment = Clockwise moment.40gf×40cm=80gf×d⇒d=40×4080⇒d=20cm40gf \times 40cm = 80gf \times d \[0.5em] \Rightarrow d = \dfrac{40 \times 40}{80}\[0.5em] \Rightarrow d = 20 cm \[0.5em]40gf×40cm=80gf×d⇒d=8040×40​⇒d=20cmTherefore the bar will be in balanced position if weight of 80gf is placed at a point of 20cm to the right of O.

Question 11

Question 13Figure shows a uniform metre rule placed on a fulcrum at its mid-point O and having a weight 40gf 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 40gf?

Accepted Answer

(i) We know that when anticlockwise moment is equal to clockwise moment then the rule will be in equilibrium.Let us check,(a) Anticlockwise moment = Fr40gf×(50−10)cm=40×40=1600gfcm40gf \times (50-10)cm = 40 \times 40 \[0.5em] = 1600 gfcm40gf×(50−10)cm=40×40=1600gfcm(b) Clockwise moment = Fr20gf×(90−50)cm=20×40=800gfcm20gf \times (90-50)cm = 20 \times 40 \[0.5em] = 800gfcm20gf×(90−50)cm=20×40=800gfcmSo, the meter rule will not be in equilibrium. It will turn in anticlockwise direction.(ii) To balance the rule in equilibrium 40gf should be kept on right side.Let distance from middle = d cmSo clockwise moment becomes800gfcm+(40gf×dcm)800 gf cm + (40gf \times d cm)800gfcm+(40gf×dcm)As we know, the principle of moments states thatAnticlockwise moment = Clockwise moment1600gfcm=800gfcm+40gf×dcm1600gfcm = 800gfcm + 40gf \times d cm1600gfcm=800gfcm+40gf×dcmTherefore,d=80040=20cmd = \dfrac{800}{40} = 20cmd=40800​=20cmSo, the additional weight should be placed at 70cm mark to bring the rule in equilibrium.

Question 12

Question 14When a boy weighing 20 kgf sits at one end of a 4m long see-saw, it gets depressed at its end. How can it be brought to the horizontal position by a man weighing 40 kgf.

Accepted Answer

As we know, the principle of moments states thatAnticlockwise moment = Clockwise moment20Kgf×2m=40Kgf×d⇒d=20×240⇒d=1m20 Kgf \times 2m = 40 Kgf \times d \[0.5em] \Rightarrow d = \dfrac{20 \times 2}{40} \[0.5em] \Rightarrow d = 1m20Kgf×2m=40Kgf×d⇒d=4020×2​⇒d=1mSo it is proved that horizontal position can be achieved if a person sits at a distance of 1m from the centre on the side opposite to the boy.

Question 13

Question 15A physical balance has its arms of length 60 cm and 40 cm. What weight kept on a pan of longer arm will balance an object of weight 100 gf kept on other pan?

Accepted Answer

As we know, the principle of moments states thatAnticlockwise moment = Clockwise moment100gf×40cm=W×60cm100gf \times 40cm = W \times 60cm100gf×40cm=W×60cmWeight on the longer panW=(100×40)60W=66.67gfW = \dfrac{(100 \times 40) }{60} \[0.5em] W = 66.67gfW=60(100×40)​W=66.67gfSo, a weight of 66.67gf kept on the pan of longer arm will balance the object of weight 100 gf kept on the other pan.

Question 14

Question 16The diagram shows a uniform meter rule weighing 100 gf, pivoted at its centre O. Two weights 150gf and 250gf hang from the point A and B respectively 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 moment, and(iv) the distance from O where a 100gf weight should be placed to balance the metre rule.

Accepted Answer

(i) The total anticlockwise moment about the centre o=150gf×40cm=6000gfcm= 150gf \times 40cm \ = 6000gf cm=150gf×40cm=6000gfcm(ii) The total clockwise moment about the centre o=250gf×20cm=5000gfcm= 250gf \times 20cm \[0.5em] = 5000gf cm \[0.5em]=250gf×20cm=5000gfcm(iii) The difference of anticlockwise and clockwise moment=6000−5000=1000gfcm= 6000 - 5000 \[0.5em] = 1000gf cm \[0.5em]=6000−5000=1000gfcm(iv) As we know, the principle of moments states thatAnticlockwise moment = Clockwise moment150gf×40cm=250gf×20cm+100gf×d6000gf=5000gf+100gf×dd=1000100d=10cm150gf \times 40cm = 250gf \times 20cm + 100gf \times d \[0.5em] 6000gf = 5000gf +100gf \times d \[0.5em] d = \dfrac {1000}{100} \[0.5em] d = 10cm150gf×40cm=250gf×20cm+100gf×d6000gf=5000gf+100gf×dd=1001000​d=10cmSo d = 10 cm on the right side of o

Question 15

Question 17A uniform metre rule of weight 10 gf is pivoted at its 0 mark.(i) What moment of force depresses the rule?(ii) How can it be made horizontal by applying a least force?

Accepted Answer

(i) Moment of force = FrSubstituting the values of F and r we get,10gf×50cm=500gfcm10gf 	imes 50cm = 500gfcm10gf×50cm=500gfcm(ii) In order to make the rule horizontal by applying least force distance has to be max as moment of force will remain same.So,F=500100=5gfF = \dfrac{500}{100} = 5gfF=100500​=5gfTherefore, a force of 5 gf upwards at 100 cm can balance the rule.

Question 16

Question 18A uniform half metre rule can be balanced at the 29.0 cm mark when a mass 20g is hung from its one end.(a) Draw a diagram of the arrangement.(b) Find the mass of the half metre rule.(c) In which direction would the balancing point shift if 20g mass is shifted inside from its one end?

Accepted Answer

(b) Suppose, if the mass of the meter rule be M.Given, uniform meter rule, so weight (Mg) will act at 25 cm.Mg produces anticlockwise moment about point o.Now, in order to balance the 20g weight is tied at 50cm point and it acts in the clockwise direction.As we know, the principle of moments states thatAnticlockwise moment = Clockwise momentMg(29−25)=0.02g(50−29)M=21×0.024=0.105kg=105gMg(29 - 25) = 0.02 g (50-29) \[0.5em] M = \dfrac{21\times 0.02}{4} \[0.5em] = 0.105 kg \[0.5em] = 105gMg(29−25)=0.02g(50−29)M=421×0.02​=0.105kg=105gThe weight of half the meter rule is 105 g.(c) The balancing point will shift towards the 25 cm mark.

Question 17

Question 19A uniform metre rule of mass 100 g is balanced on a fulcrum at mark 40 cm by suspending an unknown mass m at the mark 20 cm.(i) Find the value of m.(ii) To which side the rule will tilt if the mass m is moved to the mark 10 cm?(iii) What is the resultant moment now?(iv) How can it be balanced by another mass 50 g?

Accepted Answer

(i) As we know, the principle of moments states thatAnticlockwise moment = Clockwise moment100g×(50−40)cm=m×(40−20)cm⇒100×10=m×20⇒m=50g100g \times (50-40)cm = m \times (40-20)cm \[0.5em] \Rightarrow 100 \times 10 = m \times 20 \[0.5em] \Rightarrow m = 50g \[0.5em]100g×(50−40)cm=m×(40−20)cm⇒100×10=m×20⇒m=50g(ii) When the mass m is shifted to mark 10cm , it results in rule being shifted on the side of mass m in anticlockwise direction.(iii) Anticlockwise moment is produced when a mass of m grams is moved towards the mark of 10cm.100gf×(50−40)cm=1000gfcm100gf \times (50-40)cm = 1000gfcm100gf×(50−40)cm=1000gfcmTherefore, the resultant moment will be1500gfcm−1000gfcm=500gfcm(anticlockwise)1500gfcm - 1000 gfcm = 500gfcm (anticlockwise)1500gfcm−1000gfcm=500gfcm(anticlockwise)(iv) As we know, the principle of moments states thatAnticlockwise moment = Clockwise momentSo,100gf×(50−40)cm+50gf×d=50gf×(40−10)cm⇒1000gfcm+50gf×d=1500gfcm⇒50gfd=500gfcm⇒d=10cm100gf \times (50-40)cm + 50gf \times d = 50gf \times (40-10)cm \[0.5em] \Rightarrow 1000gfcm + 50gf \times d = 1500gfcm \[0.5em] \Rightarrow 50gfd = 500gfcm \[0.5em] \Rightarrow d = 10 cm100gf×(50−40)cm+50gf×d=50gf×(40−10)cm⇒1000gfcm+50gf×d=1500gfcm⇒50gfd=500gfcm⇒d=10cmHence, we can balance 50gm at 50cm

Question 18

Question 20In figure below, a uniform bar of length l m is supported at its ends and loaded by a weight W kgf at its middle. In equilibrium, find the reactions R1 and R2 at the ends.[Hint : In equilibrium R1 + R2 = W and R1×l2=R2×l2R_{1}	imes\dfrac{l}{2} = R_{2}	imes\dfrac{l}{2}R1​×2l​=R2​×2l​]

Accepted Answer

As we know, the principle of moments states thatAnticlockwise moment = Clockwise momentR1 + R2 = WGiven, the system is in equilibrium soR1×l2=R2×l2R_{1}	imes\dfrac{l}{2} = R_{2}	imes\dfrac{l}{2}R1​×2l​=R2​×2l​Since, R1 = R2 and 2R1 = WTherefore, R1 = W/2 = R2

Question 19

Question 3The centre of gravity of a hollow cone of height h is at distance x from its vertex where the value of x is:h3\dfrac{	ext{h}}{3}3h​h4\dfrac{	ext{h}}{4}4h​2h3\dfrac{2	ext{h}}{3}32h​3h4\dfrac{3	ext{h}}{4}43h​

Accepted Answer

2h3\dfrac{2	ext{h}}{3}32h​Reason — The centre of gravity of a hollow cone is at a height h3\dfrac{	ext{h}}{3}3h​ from the base.Hence, from the vertex the height is :h - h3\dfrac{	ext{h}}{3}3h​ = 3h-h3\dfrac{	ext{3h-h}}{3}33h-h​ = 2h3\dfrac{2	ext{h}}{3}32h​

Question 20

Question 2Draw a neat labelled diagram for a particle moving in a circular path with a constant speed. In your diagram show the direction of velocity at any instant.

Accepted Answer

The below diagram shows a particle moving in a circular path with a constant speed and its direction of velocity:

Force

Exercise 1A Long Questions

Exercise 1A Numericals

Exercise 1B Multiple Choice Type

Exercise 1C Long Questions