A girl pushed a 25 kg lawn mower. If F= 30 N and θ = 37 degrees?

Q. A girl pushed a 25 kg lawn mower as shown. If F= 30 N and θ = 37 degrees

a. What is the acceleration of the mower, and
b. What is the normal force exerted on the mower by the lawn? Ignore friction.

Answer
a)

F(horizontal) = F(applied)cos37
Note that F(horizontal)= ma
ma=F(applied)cos37 -> a = F(applied)cos37 / m; substitute in numbers

a = 30Ncos37/25kg = .958 m/s2

b)

F(vertical)=F(normal) - F(applied)sin37 - F(gravity)
Note that F(vertical) = 0 because the lawn mower is not moving up or down
0=F(normal)-F(applied)sin37-F(gravity) -> F(normal) = F(applied)sin37 + F(gravity)
Note that F(gravity) = mg (where g = 9.8m/s2); substitute in values

F(normal) = 30Nsin37 + 25kg(9.8m/s2)
F(normal) = 263 N

A gardener mows a lawn with an old-fashioned push mower.?

HT

The handle of the mower makes an angle of 36 degree with the surface of the lawn.

a. If a 208 N force is applied along the handle of the 16 kg mower, what is the normal force exerted by the lawn on the mower?

b. If the angle between the surface of the lawn and the handle of the mower is increased, does the normal force exerted by the lawn increase, decrease, or stay the same? explain

Answer
a. 208 sin (36) + 16*9.8
b. Increase, because value of sin (a) will increase..

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