Question

QUESTION FIVE [Total: 20 Marks) a. You came across a picture of three coplanar forces acting at a point, P. The first force is 200N and acting at 20° to the horizontal; the second 400N and acting at 165° to the horizontal; the third 500N acting at 250° to the horizontal. Draw what you saw in the picture using an appropriate scale. Use protractor for the angles. AP[8 marks) b. With the information in a), determine by resolution of forces i. The magnitude of the resultant of the three coplanar Forces The direction of the resultant of the three coplanar forces. AP [7 marks] c. A horizontal force with a mass of 100kg is being applied to a body to move it at a uniform velocity through a distance of 20m in 8 seconds in the direction of the force Determine the value i. The Force The Work Done The Power AP[5 marks]

please I need help

Answer 1

a IFJ-200N 16400 N 1650 74 250 IF I = SOON (6) Ę = 15il Cos 20° î + Fij sim 20° 2008 (os 20° + 200 x Sin 20 } 187:94 î + + 68.4 2 Fy - ll ( Con 165° + Sim 165) 400 (Cos 165*ê + Sim 165° 5) -386.4 +103.53 ÎN E = 153) + Sin 250 ºf 500 (Cos2root sin + Sin 25009) = -171i - 469.85 (con 250 The resaltant force F = R + F2 F3 = -369.46î - 297. 92 ŷ N

Magnitude of forece in F1 = √ (369.46)²+(297.9292 N ? 474.6N direction of the resultant forece is i FB -369:46 î - 297.92 CU 2 474.6 2 -0.718 i - 0.628 (C) Stace a taky 8 second to dating The the magnitude of Fonce is f=100 9.8 N 980 N which is applied to the be body through and displace ment is 8220m- t=8 see the worck done is for 980.20. 19600 N and is 2 power 19600 - 2450 watt 8 t