Forces in action assessed homework

 

3.2 Forces in action MS

1)      1) A

2)      2) B

3)      3)D D

4)     4)  F = m x a = 24.1 x 3.5 [1]

F = 84.4N [1]

5)     5  a = F/m = 18/0.61 [1]

a = 30ms^-2 [1]

6)      6 F = m x a = 60 x 2.3 = 138N [1]

Therefore force on 2^nd skater = 138N [1]

a = F/m = 138/55 = 2.5ms^-2 [1]

7

8)Moment = Fd = 73.1 x 0.25 [1]

Moment = 18.3Nm [1]

 

9) Anticlockwise = 450 x 1.5 = 675Nm [1]

Clockwise moment = 675 = 500d; d = 675/500 = 1.35m [1]

 

10) . (i)      1        3600 ´ 1.0 = X ´ 2.5                                                                                      C2

one mark for one correct moment, one mark for the

second correct moment and equated to first moment       

                                                                                                                               A0

2        X = 1440 (N)                                                                                                  C1

Y = 3600 – 1440        or 3600 ´ 1.5 = Y ´ 2.5                                                A1

    = 2160 (N)                                                                                                  B1

(ii)     Not a couple as forces are not equal                                                              B1

and not in opposite directions / the forces are in the same direction