Question

A force P is applied on the crate at A as shown. Determine

(a) the moment of the force P about C,

(b) the magnitude and sense of the vertical force applied at E that creates the same moment about C,

(c) the magnitude and direction (from horizontal) of the smallest force applied at E that creates the same moment about C.

If P=500 N, a=25°, a=300 mm, b=500 mm, c=350 mm, d=50 mm

a) Moment of P about C(N · m, 3 sigfig, CW+)=

b) Magnitude of Force at E(N, 3 sigfig) =

c) (i) Magnitude of Force at C(N, 3 sigfig) =

c) (ii) Direction of Force at C (Degrees, 3 sigfig, 0-180 from + ve x-axis )=

Answer 1

350 mm E 500-505 (25) = 211.309 N 500 N B 250 A E 500 x cos(25) - 453.54 N 300 mm LI 50mm temic 500mm a E Mc =0 5 -Mc - 4530154 x 300 +2 211.309 x 500=0 Mc = - 30291.7 N-mm Mc = 30.292 Nom resign indicate direction is in opposite sence as assumet, 30.292 Nom magnitude ccw (0) . Direction

P E 350 mm 150 mm TO in dulce a counter clockwise moment of magnitude apply a Vertical downa af E. 30.292 Nom we need to ) - word force o PX 150 - 30291.7 Ccw sign PE 201.945 N magnitude sense & down word direction

psina AP ha Posa 300 35291.7 mm 150 (Tiame0 psina x150 tposa *300 i 302.91.750 To have minimum value of C differentiase this aquation worot orp sin(a) *150) a(peos 2) 300) Id o dd 300 Px sina 150P. Cosa tand 150 300 a 26. 5650 #pat inequn to get value of (P px sin( 26.565)*150 + p. sos ( 26,565) +300 -30291.750 P 90.312 N