Class 12 Mathematics Statics Notes and Important Questions

Class 12 Statics Notes (Mathematics)

Mechanics UNIT 8: MECHANICS Exercise 18.1 1. Three forces acting on a particle are in equilibrium; the angle between the first and second is 90° and that between the second and third is 120°; find the ratio of the forces. Soln: Let the three forces P, Q, R acting at O are in equilibrium. Here, angle between P an Q is 90o, angle between Q and R is 120o  Angle between Q and R will be 360o – ( 90o + 120o) = 150o Using Lami's Theorem, we have P sin120° = Q sin150° = R sin90°  P

= Q

= R 1  P 3 = Q 1 = R 2  P : Q : R = 3 : 1 : 2. 2. The sides AB and AC of a triangle ABC are bisected in D and E; show that the resultant of forces represented by BE and DC is represented in magnitude and direction by 3 2 BC. Soln: Since AB and AC are bisected at D and E, so CE  = CE  2 and DB  = AB  2 ........... (i) Now, the resultant force represented by BE and DC is BE  + DC  = ( ) BC  + CE  + ( ) DB  + BC  =       BC  + CA  2 +      AB  2 + BC  [ using (i)] = 2BC  + 1 2 ( ) CA  + AB  = 2BC  + 1 2 CB  = 2BC  – 1 2 BC  = 3 2 BC  . 3. Find a point within a quadrilateral such that, if it be acted on by forces represented by the lines joining it to the angular points of the quadrilateral, it will be in equilibrium. Soln: We have, OA  + OB  + OC  + OD  = 0  ............. (i) Let E, F, G, H are the middle points of the sides AB, DC, BC and AD respectively. Then, OE  = OA  + OB  2 and OF  = OC  + OD  2 . Now, OE  + OF  = OA  + OB  2 + OC  + OD  2 = 1 2 [OA  + OB  + OC  + OD  ] = 1 2 × 0  = 0  P Q R 150 120

*374* Solution Manual to Basic Mathematics Mechanics 10 kg 16 kg x T T x A B F C D H Which shows that,OE  and OF  are equal and opposite. i.e.,OE  = – OF  Hence, O is the middle point of EF.Similarly, O is the middle point of GH.  The point of intersection of the lines joining the middle points of the opposite sides of the quadrilateral is required point. 4. The sides BC and DA of a quadrilateral ABCD are bisected in F and H respectively. Show that if two forces parallel and equal to AB and DC act on a particle, then the resultant is parallel to HF and equal to 2HF. Soln: Let ABCD is a quadrilateral and F and H are the middle points of BC and DA respectively. Now, the resultant of the forces parallel to AB and DC is given by AB  + DC  = ( ) AH  + HB  + ( ) DH  + HC  = ( ) HB  + HC  + ( ) AH  + DH  = ( ) 2HF  – ( ) HA  + HD       HB  + HC  2 = HF  = 2HF  – 0  = 2HF  [ HA  and HD  are equal and opposite.] Hence, the resultant of AB  and DC  is parallel to HF  and is equal to 2HF. 5. A heavy chain has weights of 10 kg and 16 kg. attached to its ends and hangs in equilibrium over a smooth pulley. If the greatest tension of the chain is 20 kg. wt., find the weight of the chain. Soln: Let T be the tension of the string, length of the chain be l and mass of chain be m.  Mass per unit length of chain = m l ; Mass of (l – x) length of chain =     m l (l – x) Mass of x length of chain =     m l x; Let, T be the tension of the string . At equilibriumwe have, T = 16 +     m l (l – x) and T = 10 +     m l x, since tension in the single rope is the same. Then,16 +     m l (l – x) = 10 +     m l x or, 6 =     m l (x – l + x) or, 6 =     m l (2x – l) ........ (i) It is given that greatest tension of chain is 20 kg wt. Then, 20 = 10 +     m l x  10 =     m l x ................... (ii) Dividing eqn (i) by (ii), we get

10 = 2x - l x or, 6 10 = 2 – l x or,l x = 2 – 6 10 or, l x = 14 10 or, x = 5 7 l Putting the value of x in eqn (i), we get 6 = m l     2 × 5 7 l - l or, 6 = m l     3l 7  m = 14  Mass of chain = 14 kg wt.

Statics *375* Mechanics 10 kg 16 kg A D B O C W T1 T2 30° 60° “Alternately” Let w1 be the weight of the part of the chain (from top to bottom) containing the weight of 10 kg at its end. Then, T = w1 + 10 or, 20 = w1 + 10 or, w1 = 10 kg Again, we be the weight of the part of the chain, containing the weight of 16 kg at its another end. Then, T = w2 + 16 or,20 = w2 + 16 or,w2 = 4 kg  Weight of the chain is w1 + w2 = 10 + 4 = 14 kg 6. Two men carry a weight 50N between two strings fixed to the weight; one string is inclined at 30° to the vertical and the other at 60°, find the tension of each string. Soln: Let T1 and T2 be the tensions of the strings OA and OB respectively. W = 50 N weight along OC such that AOD = 30° and BOD = 60°. Using Lami’s theorem, we have T1 sin BOC = T2 sin AOC = W sinAOB or, T1 sin 120° = T2 sin 150° = 50 sin90°  T1 = 50 1 × 3 2 = 25 3 N & T2 = 50 1 × 1 2 = 25 N. 7. A body weighing 4 N is supported by a string attached to a fixed point and is pulled from the vertical by a horizontal force of 3 N. Find the angle, the string will make with the vertical and the tension of the string. Soln: Let  is the angle made by the string OB with the vertical OA and T be the tension of the string. Since 4N is the weight of the body acting to the body horizontally pulled from the vertical by a horizontal force 3 N. Resolving the tension T horizontally and vertically, we get T cos = 4 N and T sin  = 3 N ............... (i) Squaring and adding, we get T2 = 42 + 32 or, T2 = 25  T = 5N Substituting the values of T in (i), we get 5sin = 3 or, sin = 3 5  = sin–1     3 5 . 8. A body of weight 65 N is suspended by two strings of lengths 5 m and 12 m attached to two points in the same horizontal line whose distance apart is 13 m. find the tensions of the strings. Soln: Let, AC and BC be two strings of lengths 5 m and 12 m respectively attached at A and B of the horizontal line AB = 13 m. Since (AC)2 + (BC)2 = (AB)2  Triangle ACB is right angled at C. Let the weight of 65N be acting vertically downwards along the line CE. Produce EC to meet AB at D such that CD  AB Let, CBA =  ACD = 90° – BCD = CBD =  If T1 and T2 are the tensions along the strings CA and CB, then three forces T1, T2 and 65N acting at C are in equilibrium. Using Lami's theorem, we have B A D 1 3 m C 9 0 ° +  5 c m 1 2 c m   6 5 N T1 T2

*376* Solution Manual to Basic Mathematics Mechanics A B E C 90°  0.12 m 0.05 m 0.13 cm T1 T2 D  T1 sin BCE = T2 sin ACE = 65 sin ACB or, T1 sin(90 + ) = T2 sin(180 - ) = 65 sin90° or, T1 cos = T2 sin = 65 1 T1 = 65 cos and T2 = 65 sin ............... (i) From ACB, we get cos = BC AB = 12 13 and sin = AC AB = 5

 T1 = 65 ×     12 13 = 60 N and T2 = 65 ×     5 13 = 25 N. 9. The ends of an inelastic and weightless string 0.17 m long are attached to two points 0.13 m apart in the same horizontal line and a weight of 4 N is attached to string 0.05 m form one end. Find the tension in each portion of the string. Soln: Let AC and BC be two strings of length AC = 0.05 m and BC = (0.17 – 0.005) m = 0.12 m. Here end points of the strings are attached at A and B such that AB = 0.13 m. Let the weight of 65N be acting vertically downwards along the line CE which is 0.05 m from the end A. Since (0.05)2 + (0.12)2 = (0.13)2 or, (AC)2 + (BC)2 = (AB)2  triangle ACB is right angled at C Produce EC to meet AB at D such that CD  AB. Let, CBA =  ACD = 90° – BCD = CBD =  If T1 and T2 be the tensions along the strings CA and CB respectively then three forces T1, T2 and 4N acting at C are in equilibrium. Using Lami's theorem, we have T1 sin ECB = T2 sin ECA = 4 sin ACB or, T1 sin (90 + ) = T2 sin (180 - ) = 4 sin90° or, T1 cos = T2 sin  = 4 T1 = 4cos and T2 = 4sin ........... (i) From ACB, we get cos = BC AB = 0.12 0.13 and sin  = AC AB = 0.05 0.13  T1 = 4 × 0.12 13 = 3.69 N and T2 = 4 × 0.05 0.13 = 1.54 N. 10. A uniform sphere of weight 3 N rests in contact with a smooth vertical wall. It is supported by a string whose length equals the radius of the sphere, joining a point on the surface of the sphere to a point of the wall. Calculate the tension in the string and the reaction of the wall. Soln: Let the point of contact of the wall and the sphere be A and GH be the string, where G and H are the point on the sphere and the point on the wall respectively. The weight 3N of the sphere acting at O, the tension T in the string HG and the reaction R at point A keeping the sphere in equilibrium. Here, HG = OG = OA

Statics *377* Mechanics and HAO = 90°, G is the middle point of OH. Since, middle point of hypotaneous is equidistance from each vertices, we have AG = OG = OA.  The triangle AOG is equilateral triangle and AOG = 60° Using Lami's theorem, we have T sin 90° = R sin (90° + 60°) = 3 sin (180° - 60°) or, T 1 = R

= 3

or, T = 2R = 2 3 .  T = 2 3 = 3.46 N and R = 3 = 1.73 N. 11. Forces P, Q, R acting along OA, OB, OC where O is the circumcentre of the triangle ABC, are in equilibrium, show that: i) P a cos A = Q b cos B = R c cos C ii) P a2(b2 + c2 - a2) = Q b2(c2 + a2 - b2) = R c2(a2 + b2 - c2) Soln: Let O be the circumcentre of the triangle ABC. Since the angles at centre O are double of the corresponding angles at the circumference.  BOC = 2A, COA = 2B and AOB = 2C ............ (i) i) Using Lami's Theorem, we have P sin BOC = Q sin COA = R sin AOB or, P sin 2A = Q sin 2B = R sin 2C or, P 2sinA.cosA = Q 2sinB.cosB = R 2sinC.cosC or, P (2R sinA)cosA = Q (2R sinB)cosB = R (2R sinC)cosC where R denotes the radius of circumscribed circle of the triangle ABC such that a = 2R sinA, etc.  P a cosA = Q b cosB = R c cosB ...................... (ii) ii) We have, a cosA = ab2 + c2 - a2 2bc = a2(b2 + c2 - a2) 2abc , Similarly, bcosB = b2(c2 + a2 - b2) 2abc and cosC = c2(a2 + b2 - c2) 2abc Substituting the values of acosA, bcosB, ccosC on (ii), we get P a2(b2 + c2 - a2) 2abc = Q b2(c2 + a2 - b2) 2abc = R c2(a2 + b2 - c2) 2abc  P a2(b2 + c2 - a2) = Q b2(c2 + a2 - b2) = R c2(a2 + b2 - c2)

*378* Solution Manual to Basic Mathematics Mechanics 12. O is the orthocentre of the triangle of the triangle ABC. Forces P, Q, R acting along OA, OB, OC are in equilibrium. Prove that P BC = Q CA = R AB. Soln: Let ABC be the triangle and AD, BE, CF are the perpendiculars from its vertices to the sides BC, CA and AB respectively. Then the point of intersection of them is O, called orthocentre of the triangle ABC. Now, we find BOC Since OEA = OFA = 90° and BOC = EOFEOF + EAF = 180°BOC =  – A Thus, BOC =  – A, COA =  – B, AOB =  – c ........... (i) Using Lami's theorem, we have P sin BOC = Q sin COA = R sin AOB or, P sin( - A) = Q sin( - B) = R sin ( - C) . or, P sin A = Q sin B = R sin C or, P Q R = = BC CA AB 2r 2r 2r [ Using sine law]  P BC = Q CA = R AB . 13. ABC is a triangle and D, E, F are the middle points of the sides BC, CA, AB respectively. Show that the forces represented by the straight lines AD, BE, CF acting at a point are in equilibrium. Soln: Since D, E, F are the midpoints of BC, CA and AB respectively.  AD  = AB  + AC  2 , BE  = BA  + BC  2 , CF  = CB  + CA  2 ............... (i) Now, the resultant of the forces AD  , BE  , CF  is given by AD  + BE  + CF  = AB  + AC  2 + BA  + BC  2 + CB  + CA 

= 1 2 [AB  + AC  + BA  + BC  + CB  + CA  ] = 1 2 [(AB  + BA  ) + (AC  + CA  ) + (BC  + CB  )] = 1 2 [(AB  – AB  ) + (AC  – AC  ) + (BC  – BC  )]= 1 2 [ 0  + 0  + 0  ] = 0   The forces represented by AD, BE and CF acting at O are in equilibrium. 14. A body of weight 20 N, which hangs by a string is pushed to one side by a horizontal force so that the string makes an angle of 60° with the vertical; find the horizontal force and the tension of the string. Soln: Let, T be the tension of the string makes an angle 60o with the vertical. force 20 N acing vertically downward is pushed by horizontal force F such that forces T, F, 20 N are equilibrium. Using Lami's theorem, we have F sin 120° = T sin 90° = 20 sin 150° or, F

= T 1 = 20

or, 2F 3 = T = 40  F = 20 3 N and T = 40 N.

Statics *379* Mechanics 15. A heavy chain of length 9 m and weighing 18 N has a weight of 6 N attached to one end is in equilibrium hanging over a smooth peg. What length of the chain is on each side ? Soln: Let length of chain on one side = x m Then thelength of string on other side = (9 – x) metre. Here, weight of 9 m string is 18 N then weight of 1 m string is 18 9 = 2 N here, string hanging over a smooth peg is in equilibrium  2x = 2(9 – x) + 6 or, x = 24 4 = 6 m, which is required. Hence, the length of the chain on each sides are x and (9 – x) metres i.e. 6 metres and 3 metres. Ans. 16. A uniform plane lamina in the form of a rhombus, one of whose angles is 120°, is supported by two forces applied at the centre in the directions of the diagonals so that one side of the rhombus is horizontal; show that if P and Q be the forces and P be the greater, then P2 = 3Q2. Soln: Let the uniform plane lamina in the form of rhombus be ABCD, such that BAD = BCD = 120° ABC = ADC = 60° The components of the forces P and Q in vertical direction balance the weight of lamina W acting vertically downwards. Now, resolving the forces horizontally, we have P cos 60° = Q cos 30° or, P 1 2 = Q     3 2 or, P = 3 Q .............. (i)  P > Q and P2 = 3Q2. Hint and Solution of MCQ's 1. l be the length of chain. Let w be the weight of chain such that x is the weight along the one end where 5 kg wt is attached then w – x will be the weight along other end where 8 kg weight is attached  x + 5 = 10 …… (i) (w – x) + 8 = 10 ……. (ii) from (i) x = 5 when x = 5, from (ii) w – 5 + 8 = 10 w = 7 kg wt. 2. Let  be equal angle, made by weight w acting vertically downward, to the strings attached on the weight If T1 and T2 are the tension then T1 sin (180o – ) = T2 sin( 180o – ) = w sin2  T1 = w sin sin2 = w 2cos  T2 = w sin sin2 = w 2cos  the tension of the strings are equal 3. Let  = 45o be the angles made by two strings, whose tension are T1 and T2, to the vertical line then, using (2) we get T1 = T2 = w 2cos = 20 2 cos45o = 10 2 N 180°– T1 T2 W 180°–   45° 45° 135° 135° T1 T 2 W = 20 N

*380* Solution Manual to Basic Mathematics Mechanics 4. Using Lami's theorem P sin BOC = Q Sin AOC = R sin AOB  P sin2A = Q sin2B = R sin 2C 5. Let w be the weight of chain for (12 + 3)m = 15m length.  weight for 3m chain = 3w 15 = w 5 N. weight for 12m chain = 12w 15 = 4w 5 N Here, w 5 + 6 = 4w

 w + 30 = 4w  3w = 30  w = 10N. 6. The statement is Lami's theorem. 7. If  P ,  Q ,  R , the three forces acting at a point can be represented by the sides of a triangle taken in order, then forces are in equilibrium i.e.  P +  Q +  R = 0. 8. If three forces P, Q, R acting at O along OA, OB, OC in equilibrium, then using Lami's theorem P SinBOC = Q SinCOA = R SinAOB 9. Here, three forces acting at a point are in equilibrium, so using lami's theorem P sin90o = Q sin120o = R sin150o  P 1 = Q

= R

 P: Q: R = 1: 3 2 : 1

 P: Q: R = 2: 3 : 1. O P Q A B C 150° P R Q 120° 90° 6N 3m 12 m O P R Q B A C