Physics: Newton-motion?

Question

A person pushes a 13.0 kg lawn mower at constant speed with a force of 78.0 N directed along the handle, which is at an angle of = 45.0° to the horizontal.

1. Calculate the horizontal retarding force on the mower

2. Calculate the normal force exerted vertically upward on the mower by the ground.

3. Calculate the force the person must exert on the lawn mower to accelerate it from rest to 1.2 m/s in 2.0 seconds

Answer 1

gosh....i learned about this last year....but I cant remember how to work out the formula..........

sorry

Answer 2

Constant speed means the net acceleration is zero, which means that the net force on the mower is also zero.

The forces on the mower are gravity W, the pushing force P, the drag force D, and the normal force N.

All of these are vectors in that they have direction. But for the answers, we need to know their horizontal and vertical components. W and N are vertical; D is horizontal. Since P is at an angle of 45 degrees, it needs to be split into a horizontal component Ph and a vertical component Pv.

So the net force being 0 means that the sum of the horizontal forces is 0 and the sum of the vertical forces is 0, or:

D = Ph
N = W + Pv

Since we know P and the angle, we can compute Ph and Pv. Since D is the retarding force, and D = Ph, we can answer question 1.

Since we know the mass of the lawn mower and the acceleration due to gravity (9.8 m/s) we can compute W.

Since N = W + Pv, we can compute N.

To answer 3, we need to compute the coefficient of friction C. We know:

D = C x N or C = D/N

Since we know both D and N, we can compute C.

For question 3, we need to accelerate the lawn mower. If the acceleration is A and the velocity is V, we know:

V = A x T

We know V and we know T so we can compute A.

Now we are increasing the net force so that the net acceleration = A in the horizontal direction. But the sum of the vertical forces must still be zero. So the new equations are:

M x A = Ph' - D' (horizontal forces give net acceleration)
N' = W + Pv' (vertical forces sum do not)

D' = C x N'
Ph' = K x Pv'

Where:

N' is the new normal force, which changes because P has changed.
D' is the new drag force, which has changed because P has.
Ph' and Pv' are the horizontal and vertical components of P' the new pushing force.
K is the ration of Ph/Pv. This doesn't change because it depends only on the geometry (in this case, the angle of the pushing force).
M is the mass of the lawnmower

So we have:

M x A = Ph' - D' = Ph' - C x N' = Ph' - C x (W + Pv') or

M x A = K x Pv' - C x (W + Pv')

This equation has only one unknown, Pv; we know the value of all the other symbols. So we simplify and compute Pv'.

Then we compute P' either directly from Pv' or by using:

Ph' = K x Pv'
P' = square root (Ph'^2 + Pv'^2)