182 ANSWERS. 



Pages 67-68. 



(1) Take moments about the fulcrum. The distance of this point 

 from the end carrying the mass 12 is (a) 3 T 6 ft. ; (b) 3^ ft. 



(2) (a) A = i 2 | tons, = nj- tons; (V) i8, i8|. 



(3) (a) P= W-, (b) P= (i 4- VI) JF. 



(6) (<*) 19.4 tons and 21.1 tons; (b) 30.5, 9.9. 



(8) Let be the angle subtended at the centre by the side 12, and 

 the angle at which the diagonal 13 is inclined to the horizon; then 



j^7 }y 



tan 6 = ^ __ ^ esc a -f cot a. 



(9) x = F z l sin 2 /(^i sin a x + /^ 2 sin a 2 ) . 



Page 75. 



(i) C=i, D = 

 = 4.2, EF= 8.9 ; reaction at A = 4.5, at F= 8.9. 



(2) H= 69.4, T= 73.8. 



(3) * = 



Pages 90-92. 



(1) T 7.68, ^ = 9.76, ^= 1 2.80 pounds. 



(2) T= 2mW, A= V4 ni 1 2 m -\- i W^ where w = r//. 



(3) The three forces W, T, A must pass through a point; cos< 

 = 2V|(i -^ 2 ), where m = l/&; T= 



(4) r= 



(5) A x = B = 



(6) ^ sr = 



the thrust ^ z in this case is to that in Ex. (5) as sin 2 a is to i. 



(7) A = W, C = D = (l/a}cosOW. 



(8) x = am, A = V^^T W, C=m W, where m = (//)*. 



(9) ^ = 1(30/4- W) tan , ^=(30/4- ^)Vitan 2 + i. 

 (10) cos0 = \(m + Vw 2 -f 32), where w = I/a. 



