Page 22 - Physics Form 5 KSSM_Neat
P. 22

Example         3

           Figure 1.10 shows a trolley of mass 1.2 kg being pulled on a table   Trolley
           by a load through a pulley. The trolley moves with an acceleration
           of 4.0 m s  against a friction of 6.0 N.
                    –2
           (a)  Sketch the free body diagram of the trolley and the load.
               Use  W = the weight of the trolley, R = normal reaction on
             KEMENTERIAN PENDIDIKAN MALAYSIA
               the trolley, F  = friction, T = tension of the string and                     Load
                          R
               B = the weight of the load.                                       Figure 1.10
           (b)  Compare the weight of the trolley, W with the normal   Note
               reaction, R.
           (c)   Calculate the resultant force acting on the trolley, F.  Based on Figure 1.11, the two forces, T
           (d)  Calculate the tension in the string pulling the trolley, T.  (the action and the reaction) are acting
           (e)   What is the mass of the load, m?                   along the string between the trolley and
               [Gravitational acceleration, g = 9.81 m s ]          the load. Since the trolley and the load are
                                                   –2
                                                                    connected by a string, both will move with
              Solution                                              the same acceleration.

           (a)  Figure 1.11 shows the free body diagram  (d)  Resultant force, F = 4.8 N
               of the trolley and the load.              Friction, F  = 6.0 N
                                                                  R
                             R                             F = T – F , thus T = F + F
                                                                    R
                                   T                       T = 4.8 + 6.0          R
                    F
                     R
                                          T                  = 10.8 N
                                                     (e)  Acceleration of the load, a = 4.0 m s –2
                          W                              Gravitational acceleration, g = 9.81 m s –2
                                          B               F  = ma               F = B – T
                           Figure 1.11                      = m × 4.0          4m = 9.81m – 10.8
           (b)  Weight of trolley, W = normal reaction, R       = 4m         5.81m = 10.8
           (c)  Mass of trolley, m = 1.2 kg               B = mg                m =  10.8
               Acceleration of the trolley, a = 4.0 m s         = m × 9.81          5.81
                                                 –2
               F = ma                                       = 9.81m               = 1.86 kg
                 = 1.2 × 4.0
                 = 4.8 N
           Formative Practice               1.1


              1.  Determine the magnitude and direction of the resultant force in the following situations.
                (a)                   9 N                (b)  5 N   2 N


                    11 N
                                             17 N
                                                                   13 N
              2.  Figure 1.12 shows the forces acting on a ball that is kicked   180 N
                simultaneously by two players.
                (a)  Sketch a diagram that shows the 240 N force, the 180 N force and     240 N
                    the resultant force.
                (b)  Calculate the magnitude of the resultant force on the ball.    Figure 1.12
                (c)  State the direction of motion of the ball.
             12                                                                         LS   1.1.4
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