Question

A child pulls on a wagon of mass 47 kg at an angle 33 degrees above the horizontal with a force of F. If the wagon accelerates at 3.7 m/s2 horizontally, what is the magnitude of the force F (in unit of N)? Assume there is coefficient of friction is 0.33, and please use g = 10 m/s2 to simplify computation.

Answer 1

Vertical forces acting on the wagon are:

1.due to earth's gravity, mg downwards (where m is mass of wagon and g is acceleration due to gravity)

2.Normal reaction, N upwards

3.Due to force applied by child, Fsin upwards, where F is the magnitude of the force applied by the child and is the angle that the force makes with horizontal. (note that sin has been multiplied to get the vertical component of force applied by the child)

Since the wagon is not accelerating vertically, upward forces must balance downward forces.

So, Fsin+N=mg=>N=mg- Fsin....................................................equation 1

Assuming acceleration towards right,horizontal forces acting on the wagon include:

1.due to friction, *N towards the left as friction acts opposite to relative motion. (Here,= coefficient of friction and N is normal reaction)

2. Due to force applied by the child, Fcos towards the right, where F is the magnitude of the force applied by the child and is the angle that the force makes with horizontal. (note that cos has been multiplied to get the horizontal component of force applied by the child)

Using newton's second law of motion, net force= mass*acceleration, we get Fcos-*N=ma where a is the acceleration.

Using equation 1 to substitute value of N, we get, Fcos-*(mg-Fsin)=ma => Fcos+Fsin=ma+mg

Here, m=47kg, g=10 m/s² ,=0.33,a=3.7 m/s²,=33 degrees.

So, Fcos33+0.33Fsin33=47*3.7+0.33*47*10=> F(cos33+0.33sin33)=47*3.7+0.33*47*10   =>F=(47*3.7+0.33*47*10)/(cos33+0.33*sin33)=323.06 N