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Solutions Complete solutions to Miscellaneous Exercise. The unit vector, u, can be obtained by using (.5. u= ( 5i+ 7j+ 5 + 7 + 5 7 = ( 5i + 7j+ = i+ j+ 8 8 8 8. (i We have ( 3 ( 6 a+ c= i+ j + i j = i+ 3 j+ 6 = i+ j 7 (ii Similarly a+ b+ c+ d= i+ 3j + 7i+ + i j 6 + i j ( (iii ab = ( i+ 3j ( 7i+ = ( 7 + ( 3 + ( = = + 7 + i+ 3+ j+ + 6 = i+ j 6 (iv Since b a = a b = (v We first obtain b + c and then find the scalar product of a and b + c : b+ c= 7i+ + i j 6 = 7 i+ j+ 6 = 6i j+ 6 ( ( ( ( ( ( a b+ c = i+ 3j 6i j+ 6 = 6 + 3 + 6 = 3 (vi The unit vector, a ˆ, in the direction of a can be found by using (.5. aˆ = ( i+ 3j = ( i+ 3j + 3 + 4 ( Next we find the scalar product c d cd = i j 6 i j = + + 66= 55 ( ( Hence we have ( cda 55 55 3 55 ˆ = 55 ( 3 9.4 44. 4.7 4 i + j = + = + 4 i 4 j 4 i j (vii We have already found a + b + c + d in part (ii, so the magnitude is ( a+ b+ c+ d = i+ j = + + = 6 6.9 3. Let θ be the angle between v and a. We use cos( θ = va (* va Evaluating each of the components of (*: va = 4i 8j+ 4 i+ j+ 4 = 8 8+ 6= ( ( ( ( v = 4i 8j+ 4 = 4 + 8 + 4 = 96 a = i+ j+ 4 = + + 4 = Substituting these into (* gives cos( θ =, taing inverse cos θ ( = cos = 9 The velocity vector is perpendicular to the acceleration vector. (.5 u = a + b + c ( ai + bj + c

Solutions 4. Similar to solution 3. Use (* of solution 3: v a= 6i+ j 3i+ 7j = 6 3 + 7 = 3 ( v = i+ j = + = 6 6 4 a = i+ j = + = 3 7 3 7 58 Substituting these into (* 3-3 cos ( θ = which gives θ= cos = 48.37 4 58 4 58 5. We first place each force into polar form: 8 ( 45 and 7.6 5 Converting these forces into i and j components via calculator 8 ( 45 =.73i.73j 7.6 5 = 6.58i + 3.8j and adding gives the resultant force R as R = (.73 6.58 i+ (.73+ 3.8 j = 6.5i 8.93j 55.46 =.84 ( 55.46.84N 6. Similar to solution 5. Putting each force into polar form 5 5, 7.8 ( 45 and 3.6( 9 and then using a calculator gives the resultant force R as R = 5.34 8.9 5.34N R 8. 9 7. y using trigonometry we have F = 39.5cos( 53 x = 3.77 N (horizontal F = 39.5sin ( 53 y = 3.55 N (vertically down 8. dding each component i, j and separately gives: (a R = ( 3 + 6 + 3i + ( + 64 + 8j + ( 7 + +9 + = 63i + 69j + 48 (b R = ( +6 + 47 + i + ( + 99 + 84 + 4j + ( +9 + 88 + 3 = 86i + 5j + 9. Using (.5 we have wor done = ( i + j + 3 ( 5i + j = 5 + = 35J. We have O = i + 3j + 7 and O = 7i j + 37 Thus O = O O = O O = 7i j+ 37 i+ 3j+ 7 = 5i 4j+ 3 ( ( (.5 wor done = F r

Solutions 3 pplying (.5 gives wor done = i 7j+ 5i 4j+ 3 ( ( ( ( ( ( = 5 + 7 4 + 3 = 468 J. (a The position vectors in i, j and components of and are y (.5 we have wor done (b Similarly we have O = i + 7 j+, O = 3i+ 4j O ( ( = O O = 3i + 4j i+ 7j+ = i 3j = ( 3i j ( i 3j = 6 + 3 + 4 = 3J. O = 3i + 4j+ 3, O = 6i+ 7j 7 = O O = 6i+ 7j 7 3i+ 4j+ 3 = 3i+ 3j ( ( y (.5 wor done = i+ j 7 3i+ 3j = 6 + 3+ 7 = 49J (. We have = O O O = ( 3i + j+ 8 ( i 3j+ 7 F = i+ 4j+ Let M be the moment then by (.5 i j M = ( i+ 4j+ ( 5i j = det 4 5 4 4 = i det j det + det 5 5 = i ( 4+ j( 5 + ( = 3i+ 6j The magnitude of M is M = 3i+ 6j = 3 + 6 + =.5 N m 3. We have O ( ( = i + j + 3, O = 5i + j = O O = 5i+ j i+ j+ 3 = ( 5+ i + ( j+ ( 3 = 6i j 5 O ( ( The unit vector, u, in the direction of (.5 Wor done = F r (.5 Moment = r F is obtained by using:

Solutions 4 Thus u= i j = 6 ( ( 6 + + 5 ( 6 5 ( 6i j 5 39 6 5 9.7 4.95 4.77 N 6 F= ( i j = ( i j 4. We need to find F r + F r + F r 3 3 ( Each vector product is evaluated by using (.4: F r = ( i+ 3j 6 ( i+ j+ i j 3 6 6 3 = det 3 6 = i det j det + det = i 8 j 4 + 4 = 8i 4j 4 Similarly F r = 7i+ j+ 5 3i j ( ( i j 5 7 5 7 = det 7 5 = i det j det + det 3 3 3 = i 8 j 8 + 8 = 8i+ 8j+ 8 lso F3 r3 = j+ 3 i+ j ( ( i j 3 3 = det 3 = i det j det + det = i( 6 j( 6 + ( 4 = 6i+ 6j 4 Substituting these into ( : 8i 4j 4 + 8i+ 8j+ 8 + 6i+ 6j 4 = i+ j+ = 5. We have by (.4 i j r s= ( ai+ b ( ci+ d = det a b c d b a b a = i det j det + det d c d c = j + = ( ad cb j( cb ad i j a + b + c d + e + f = a b c d e f (.4 ( i j ( i j det

Solutions 5 6. Since the vectors are 9 to each other we have cos( 9 = = ab ab Thus the numerator, a b, must be zero. ab = i 3j+ x 3i+ j+ x = 6 3+ x = ( ( Rearranging and solving the quadratic x 9= = 9 x = 9 = 3, 3 The vectors a and b are 9 to each other if x = 3 or x = 3. 7. We need to write the exact value of cos( 45 : cos( 45 = We have ab = ab Rearranging gives ( x ab = ab (* We evaluate each component of (* ab = i+ j+ xi+ j = x+ = x+ ( ( Substituting these into (* gives Squaring both sides Rearranging Hence a = i+ j+ = + + = 3 ( b = xi+ j = x + + = x + ( x x + = 3 + ( x+ = ( x + ( x x x 9 4 + 4 + = 9 + 8 + + = + 8x 8x 9x 9x 8x 8x+ 8 = x 8x+ 6= ( x 8 4 = gives x=4 x = 4 gives an angle of 45 between the two vectors.

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