5.8.IDENTIFY:Apply Newton’s 1st law to the wrecking ball. Each cable exerts a force on the ball, directed along the cable.SET UP:The force diagram for the wrecking ball is sketched in Figure 5.8. Figure 5.8 EXECUTE:(a)yyFma=∑cos400BTmg° −=24(4090 kg)(9.80 m/s)5.2310Ncos40cos40BmgT===×°°(b)xxFma=∑sin400BATT° −=4sin403.3610NABTT=° =×EVALUATE:If the angle 40is replaces by (cable Bis vertical), then and °0°BTmg=0.AT=5.14.IDENTIFY:Apply m=∑FaGGto each block. 0a=. SET UP:Take perpendicular to the incline and y+x+parallel to the incline. EXECUTE:The free-body diagrams for each block, Aand B, are given in Figure 5.14. (a)For B, xxFma=∑gives 1sin0Twα−=and 1sinTwα=. (b) For block A, xxFma=∑gives 12sin0TTwα−−=and 22sinTwα=. (c) yyFma=∑for each block gives cosABnnwα==. (d)For 0α→, and . For , 120TT=→ABnnw=→90α→°1Tw=, and . 22T=w0ABnn==EVALUATE:The two tensions are different but the two normal forces are the same. Figure 5.14a, b 5.21.IDENTIFY:Apply m∑F =aGGto each block. Each block has the same magnitude of acceleration a.
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