Question

The figure shows an overhead view of a 0.02 kg lemon half and two of the three horizontal forces that act on it as it is on a frictionless table. Force 1 has a magnitude of 8 N and is at 62 = 31°. Force F2 has a magnitude of 8 N and is at 82 = 32º. In unit-vector notation, what is the third force if the lemon half (a) is stationary, (b) has the constant velocity 111 – 13)) m/s, and (c) has the = 11ti m/s2, where t is time? 13-1 # – 121) 07 X 82 (a) Number j Units + (b) Number j Units i + (c) Number j Units i +

Answer 1

ng Di Fax ta fia y Esey 12 ) = 80, 0,= 310 122) 28N O2 = 32" Here Fix = F Cosoi = 6,857N Fy F, Sin ero (31) = 4.120 Fan Fa Singa 4.239 N Fay Fa cos32 = 6,784N ca) M = 0.02 kg The en dol half lemon is Stationary & Hence on denon 011 the force acting on borf is zero. along a-axis Fix & Fax + Finx = 0 - - 6.857 +4.239 + Fore = 0 F3n = 2.617 i N 7

Fory-axis Fig + fay & F3y = 0 >> 4.12 - 6.184 + Fzy=0 Ezy = 2.664 ) N so F₂ = 2.617 i + 2,664 } (b) Here the constant velocity = (11î - 13)) m/s For newtons Second tau ma= Fit24 F3 = m (die) = F, + F2 + F tience here constant verocity is question so according to do de so which is same as again fitfa tf = 0 part (a) so F₂ = 2.617 9 + 2.664 i