Question

The two blocks used in a measuring device have negligible weight. The spring is compressed 4 in. when in the position shown. The end of the screw is smooth and the coefficient of static friction at all other points of contact is p = 0.24. (Figure 1)

Answer 1

N₂ Costo & Ms N₂ Cosso - N3=0 N₂ Cos 60t Me Nz (os 30 N₂ Sinto t Ms N3 Given data ;- 4 in Spring con Steint (k)= '2015lin compression (x) = Coettícient of friction (M) = 0.24 blocks AIB: Free body diagram of MN2 มี, I kz IM N3 60 -A-TIME MNI N2 For Block A : E Fx = 0 3 E Fy=0 N2 sinto + Ms (Ma cos 60 + Ms M2 Cos38)-MsN₂ sin 38 = kx N2 [ sin 6ô + Ms (cos bô + Mis Cosso) - Ms Sin337 =kx ka so N2 = Sin 6°+ Ms ( cos 60 +420S30 - finzė) 2004 t.l (0-24) [ + (0-4) 2-6) 2 80 N2 02 / + 53 (012432 2

Mi Cos 45 - Ms N, Cos45 - N₂ (Osto- M₂N₂ (os 30 fp=0 680 80 N2 87-345-15 0591591 0.91591 For Block B - O EF=0 E Fy= N, sin 45 4Ms N, sin45°-N₂ sin 60 + M Na Sin 38=0 Ni (sin45) (1+M8] - Na sinto Ms Na singo Na Sinto-ris Na Sinzo Ni = Sin45'( 1+ M) (871345 x 3)-[(0-24) (87.245) (+) 1 / (1+0.24) 87,345 0129) 53 = a 1:24 12 650162 =74.318 IS Ni 0.8768 From PEHA, 20545'- N, Cor 45° +Nz (056* + Mg N, (0538 Ni C0545*( 48 -1) + N2 [cos 60+ Ms. Cos 30] = 791316 yilda) (6-24-1) + 871345[+618+)$ ] + 39.939 + 61.827 P = 21.888 lb