Question

Determine the magnitude of the resultant force acting on the plate and its direction, measured counterclockwise from the positive x-axis.

F₁ = 900N, F₃ = 650N,  F₂ = 750N

Answer 1

balution to the above problem: given; Diagram 12 = 750N who os 1.0 = 45° at - - - But F3 = "LON F= 650 Tan O = opposite side a - Adjacent side 10 = 36-8698° and into horizontal diagram Now balue the vertical components fasin 36-8698° F2 cos 45 -- > 12 162 9. F2 sin 45 Fi F3 cos 36. 8698

NOW Efn → Fit F2 COS 45° + FB COS 36. 8698° =0 900 + 750 cos 45º + 650 Cos 36.8692 Now & Fa - Fit Fr Cosus & F3 Los 36. 86980 > 900 + 750 COS 45° + 650 COS 36.8698° => 1950.3307 Now {Fy - F2 sin 45 - F3 Sin 36.8698 → 750 Sin 45 - 650 Sin 36.8699 -) 140. 3309 N

Now The Resultant Force is equal to R- Jizra)(Ery)? - | 1950.2001)+ (140-3309) 2 - 1955.3727 N Tan OR = Sfa sfy Tan OR = with yospect to X cous Efa 140.3309 1950.3307 FOR=4.11547 Op = 4.11547" which mean from posítve x avis, measured 4.11547 counter clockucise the resultant 1955- 3727 is acting such that it is acting North -east North east nom east east west soun

R=1955 31270 14-1157-0 Since sfx and Efy acting in having the value onese should be act in Ist Quadrant , It means Resultant is acting in St Quadrant.

Answer 2