A 960 kg car is pulling a 310 kg trailer. Together the car and trailer move forw

Question

A 960 kg car is pulling a 310 kg trailer. Together the car and trailer move forward with an acceleration of 2.00 m/s2. Ignore any frictional force of air drag on the car and all frictional forces on the trailer.
(a) Determine the net force on the car.
magnitude

N
direction


(b) Determine the net force on the trailer.
magnitude

N
direction


(c) Determine the force exerted by the trailer on the car.
magnitude

N
direction


(d) Determine the resultant force exerted by the car on the road. (Assume that the forward direction is along the +x-direction.)
magnitude

N
direction

Explanation / Answer

             The mass of the car, M = 960 kg       The mass of the  trailer, m = 310 kg       The acceleration of the car and trailer, a = 2.00 m/s2 a)
      The net force on the car is the force necessary to pull the car is
                      Fc= Ma                           = (960 kg)(2.00 m/s2)                           = 1920 N       Direction: Along positive direction __________________________________________________________ __________________________________________________________

b)       The net force on the trailer is
                         Ft = ma                              = (310 kg)(2.00 m/s2)                              = 620 N       Direction: Positive direction __________________________________________________________ __________________________________________________________ c)       The force exerted on the trailer by the car is the same force necessary       to pull the trailer since every action has an equal and opposite force.                      Ft= 620 N       Direction: Negative direction    (Since, it is in the opposite direction) ____________________________________________________________ ____________________________________________________________ d)       The net force to pull the car is 1920 N.       The force of gravity is                Fg = Mg                     = (960 kg)(9.8 m/s2)                     = 9408 N (downward)       So the resultant force the car exerts on the road is                 F = (19202+94082)                     = 9601.92 N         Direction:                = tan-1[Fg/Fc]                   = tan-1[9408 N/1920 N]                   = 78.46o    below the horizontal from the forward direction NOTE:       This is in the fourth quadrant, so the another angle is                 = 360o - 78.46o                      = 281.54o                      = 281.54o    

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