Selected Problems from Chapter 5. 1) mg T ma F=ma mg-T=ma T=m(g-a) T=700(9.8-3.8)= 4200 = 4.2 kN up.

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Presentation transcript:

Selected Problems from Chapter 5

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

2) F=20 N mg Na Na FxFx FyFy

3) F-N AB =m A a N AB =F-m A a=36-4x1.5=30 N A 36N N AB M=m A +m B =24 kg F=Ma a=F/M=36/24=1.5 m/s 2 A+B 36N

4) (consider A,B,C as one object M=3m) F=3ma hence, a=F/3m F C T1T1 a F-T 1 =ma (for block C) T 1 =F-ma=F-m(F/3m)=2F/3 A,B&C F a

5) a a a T m2gm2g a T m1gm1g T-m 1 g=m 1 a (1) m 2 g-T=m 2 a (2) add (1) & (2): (m 2 -m 1 )g=(m 1 +m 2 )a a = (m 2 -m 1 )g/(m 1 +m 2 )=(2/6)x9.80= 3.27 m/s 2

6) N a F=40 N mg 30 o y-axis mg 30 o mgsin30 o N F=40 N x-axis mgcos30 o a Along the x-axis: mgsin30 o -F=ma m(gsin30 o -a)=F m=F/(gsin30 o -a)=40/(9.80x )=14 kg

7) a a T W 1 =m 1 g 30 o a m2gm2g T a N a N WyWy T WxWx T-W x =m 1 a (1) W x =W 1 sin30 o W y =W 1 cos30 o m 2 g-T=m 2 a (2) add (1) & (2): and substitute to get the answer: a = 0.69 m/s 2 down

Continued……