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Selected Problems from Chapter 5

Selected Problems from Chapter 5. 1). T. ma. mg. F=ma mg-T=ma T=m(g-a) T=700(9.8-3.8)= 4200 = 4.2 kN up. 2). N. a. N. a. F x. F y. F=20 N. mg. mg. 3). 36N. A+B. 36N. N AB. A. M=m A +m B =24 kg F=Ma a=F/M=36/24=1.5 m/s 2. F-N AB =m A a N AB =F-m A a=36-4x1.5=30 N. 4).

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Selected Problems from Chapter 5

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  1. Selected Problems from Chapter 5

  2. 1) T ma mg F=ma mg-T=ma T=m(g-a) T=700(9.8-3.8)= 4200 = 4.2 kN up

  3. 2) N a N a Fx Fy F=20 N mg mg

  4. 3) 36N A+B 36N NAB A M=mA+mB=24 kg F=Ma a=F/M=36/24=1.5 m/s2 F-NAB=mAa NAB=F-mAa=36-4x1.5=30 N

  5. 4) a a F C F T1 A,B&C F-T1=ma (for block C) T1=F-ma=F-m(F/3m)=2F/3 (consider A,B,C as one object M=3m) F=3ma hence, a=F/3m

  6. 5) T T a a a a m1g m2g T-m1g=m1a (1) m2g-T=m2a (2) add (1) & (2): (m2-m1)g=(m1+m2)a a =(m2-m1)g/(m1+m2)=(2/6)x9.80= 3.27 m/s2

  7. 6) y-axis N a F=40 N F=40 N a x-axis mgsin30o mgcos30o 30o 30o 30o Along the x-axis: mgsin30o-F=ma m(gsin30o-a)=F m=F/(gsin30o-a)=40/(9.80x0.5-2.0)=14 kg N mg mg

  8. 7) a a a a N N T T a 30o T W1=m1g m2g Wx m2g-T=m2a (2) T-Wx=m1a (1) Wy Wx=W1sin30o Wy=W1cos30o add (1) & (2): and substitute to get the answer: a = 0.69 m/s2down

  9. Continued……

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