1 Chapter 5, problem 12 F1F1 m F 2 =? a. 2 F net =ma F1F1 m F net F 2 =?

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Presentation transcript:

1 Chapter 5, problem 12 F1F1 m F 2 =? a

2 F net =ma F1F1 m F net F 2 =?

3 -F 1 F1F1 m F net F net -F 1 F 2 =? -F 1 F net =ma

4 F net -F 1 F1F1 m F net F net -F 1 -F 1 F net =ma

5 F 2 =F net -F 1 F1F1 m F net F net -F 1 F net =ma -F 1

6 F 2 =F net -F 1 F1F1 m F net F net -F 1 F net =ma -F 1

7 F net =f net,x I+f net,y J F1F1 m F net F net -F 1 F net =ma -F 1 F 2 =F net -F 1 f net,x f net,y

8 f 2,y F1F1 m F net F net -F 1 F net =ma -F 1 F 2 =F net -F 1 f 2,y =f net,y f net,x f net,y f 2,y

9 f 2,x F1F1 m F net F net =ma -F 1 F 2 =F net -F 1 f 2,x =f net,x -f 1 f 2,y =f net,y f net,x f net,y f 2,y f 2,x

10 f 2,x, f 2,y f 2,x =f net,x -f 1 f 2,y =f net,y f 2,y f 2,x

11 f net,x =f net cosα F 2 =F net -F 1 f 2,x =f net,x -f 1 f 2,y =f net,y f net,x =f net cosα f net,y =f net sinα f 2,y f 2,x |F net |=f net α

12 f 2,x =f net cosα-f 1 F 2 =F net -F 1 f 2,x =f net cosα-f 1 f 2,y =f net sinα f net,x =f net cosα f net,y =f net sinα f 2,y f 2,x |F net |=f net α

13 |F 2 |=f 2 =((f net cosα-f 1 ) 2 +(f net sinα) 2 ) 1/2 F 2 =F net -F 1 f 2,x =f net cosα - f 1 f 2,y =f net sinα f net,x =f net cosα f net,y =f net sinα f 2,y f 2,x |F net |=f net α |F 2 |=f 2 =((f net cosα-f 1 ) 2 +(f net sinα) 2 ) 1/2

14 |F 2 |=f 2 =((ma cosα – f 1 ) 2 +(ma sinα) 2 ) 1/2 F 2 =F net -F 1 f 2,x =ma cosα - f 1 f 2,y =ma sinα f net,x =f net cosα f net,y =f net sinα f 2,y f 2,x |F net |=f net α |F 2 |=f 2 =((ma cosα – f 1 ) 2 +(ma sinα) 2 ) 1/2

15 f 2, f 2,x, f 2,y f 2,x =ma cosα - f 1 f 2,y =ma sinα f 2 =((ma cosα – f 1 ) 2 +(ma sinα) 2 ) 1/2

16 f 2, f 2,x, f 2,y f 2,x =ma cosα - f 1 f 2,y =ma sinα f 2 =((ma cosα – f 1 ) 2 +(ma sinα) 2 ) 1/2 f 2,x =1.5kg 7.3m/s 2 cos30 o - 6.8N f 2,y =1.5kg 7.3m/s 2 sin30 o f 2 =  (1.5kg 7.3m/s 2 cos30 o - 6.8N)2+ +(1.5kg 7.3m/s 2 sin30 o )2)  1/2

17 f 2,x f 2,x =ma cosα - f 1 f 2,x =1.5kg 7.3m/s 2 cos30 o - 6.8N 1.5kg(7.3(m/s^2))cos(30degree)-6.8N F 2,x = N  2.7N

18 f 2,y f 2,y =ma sinα f 2,y =1.5kg 7.3m/s 2 sin30 o 1.5kg (7.3m/s^2) sin(30degree) f 2,y = N  5.5N

19 f2f2 f 2 =((ma cosα – f 1 ) 2 +(ma sinα) 2 ) 1/2 f 2 =  (1.5kg 7.3m/s 2 cos30 o - 6.8N) 2 + +(1.5kg 7.3m/s 2 sin30 o ) 2 )  1/2 sqrt((1.5kg(7.3(m/s^2))cos(30degree)-6.8N)^2+(1.5kg (7.3m/s^2) sin(30degree))^2) f2= N  6.1N

20 f 2,x  2.7N, f 2,y  5.5N, f 2  6.1N f 2,x  2.7N f 2,y  5.5N |F 2 |  6.1N