Question

b с B a Jw E D d A crate of mass 125 kg is held in the position shown. Determine If a = 575 mm, b = c = 725 mm, and d = 900 mm, (a) the moment produced by the weight W of the crate about E, (b) the (i) magnitude and (ii) direction of the smallest force applied at A that creates moment of equal magnitude and opposite sense about E, (c) the (i) point of application, (ii) magnitude, and sense on the bottom of the crate of the smallest vertical force that creates a moment of equal magnitude and opposite sense about E. a) Moment of the Weight about E (N.m, 3 sigfig, CW+) = b) (i) Magnitude of the force at A (N, 3 sigfig) = b) (ii) Direction of the force at A (Degrees, 3 sigfig, 0-180) = c) (i) Point of Application from E (m, 3 sigfig)= c) (ii) Magnitude of the force at E (N, 3 sigtig) =

Answer 1

b С. 9 B a tw C E Mass of crate = 125 125 kg a=575mm b=c=725mm and d= goom ) solution Mass of crate : 125 kg Weight of crate Massx acceleration due to gravity (9) acceleration due to gravity (g) = 125 kg + 1* 9.81 (mis :) 9.81 mis Weight of crate - 1226-25 N Moment : a) moment produced by the weight W of crate about E. weight of crate (W) X distance between point E and point of application of weight. of point of application from point es doc - (900 - 725) mm - 175mm = 1226-25 NX 175 mm - 214593.75 N.mm Moment - 214.594 N-m (clock wise) [: 1000 mm =im] CS Scanned with CamScanner

If we b)ci) Magnitude and direction of smallest force applied at a that create a moment of equal magnitude and opposite sence. • apply force at A in vertical direction than distance between point of application of force and point E = 725+725-900 550 mm And If we apply force at A in Horizontal direction than distance between point of application of force and point E = Q = 575 mm SO for same moment smallest since 575mm > 550mm force is requried in Horizontal direction, () Magnitude of force FX 575 mm - 214.594 (Nm) Fa 214.594 (nom) [: 1000mm Im] 0.575 (m) 373 207 N in Horizontal direction ciis Direction of force B. F: 373.207 N. w с D Scanned with CamScanner

C) i) Point of application clis magnitude and sense on the bottom of the crate of the smallest vertical force that creates of equal magnitude and opposite sense about E. c) point of application from E For same moment smallest force is required where distance is maximum from point of application of force- Distance of point I from E= d= goomm Distance of points from E = 725mm 4725mm -goomm assomm goomm > ssomm. so point of application in vertical direction at bottom at point D. (ii) Magnitude of force FX goomm = 214.594 (Nm) FP 214.594 (N-m) goo mm) 214.594 (N-m) 0.9 cm) F = 238.438 N 4 upward F. b с w d D 238.438 N CS Scanned with CamScanner