Physics- Angle Forces

Question

Joe pushes down the length of the handle of an 11.1 kg lawn spreader. The handle makes an angle of 42.7 ◦ with the horizontal. Joe wishes to accelerate the spreader from rest to 1.33 m/s in 1.3 s. What force must Joe apply to the handle? Answer in units of N.

Answer 1

To accomplish that acceleration, the required horizontal force F_h is given by

F_h = a.m

where m = 11.1kg is the mass of the spreader,

and a is the desired acceleration.

The acceleration a can be obtained from its definition of

v = a.t

where t = 1.3s is time interval,

and v = 1.33m/s is velocity at the end of that time interval.

Therefore

a = v/t = 1.023m/s²

From that

F_h = 1.023m/s² x 11.1 kg = 11.356N

The question does not ask about the horizontal force F_h,

but the force F down the length of the handle.

That force F is the vector sum of the horizontal force F_h

and a vertical force F_v (which we do not need).

F_h = F.cos(42.7◦ )

Therefore

F = F_h/cos(42.7◦ ) = 11.356N / 0.7349 = 15.45N