A 980 kg car is pulling a 315 kg trailer. Together the car and trailer move forw
ID: 1328360 • Letter: A
Question
A 980 kg car is pulling a 315 kg trailer. Together the car and trailer move forward with an acceleration of 2.15 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.
(b) Determine the net force on the trailer.
(c) Determine the force exerted by the trailer on the car.
(d) Determine the resultant force exerted by the car on the road. (Assume that the forward direction is along the +x-direction.)
Explanation / Answer
car mass m1=980 kg
trailer mass m2=315 kg
acceleration a=2.15 m/sec^2
a)
the net force on the car
F=m1*a
=980*2.15
=2107 N
b)
the net force on the trailer
F=m2*a
=315*2.15
=677.25 N
c)
force exerted by the trailer on the car is equals to net force on the trailer
F=m2*a
=315*2.15
=677.25 N
d)
total mass M=(315+910)=1225 kg
the horizontal force exerted by the car on the road is,
Fx=M*a
=1225*2.15
=2633.75 N
and
weight of the car W=Fy=m1*g
=910*9.8
=8918 N
now,
resultant force exerted by the car on the road
Fnet=sqrt(Fx^2+Fy^2)
=sqrt(2633.75^2+8918^2)
=9298.78 N
and
direction is,
tan(theta)=(Fy/Fx)
tan(theta)=(8918/2633.75)
===> theta=73.54 degrees
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