Question

A boy pushes a lawn mower (m  17.9 kg) star ting from rest across a horiz ontal
lawn by applying a force of 32.9 N straight along the handle, which is inclined
at an angle of 35.1° above the horizontal. The magnitude of the mower’s acceleration
is 1.37m/s 2, which lasts for 0.58 s, after which the mower moves at a
constant velocity. Determine the magnitude of
(a) the normal force on the mower
(b) the frictional force on the mower
(c) the maximum velocity of the mower
(d) the force applied by the boy needed to maintain the constant velocity

Answer 1

a) Assuming it means find the normal force while the mower isaccelerating (and while the kid is still exerting 32.9N offorce):

Normal force (F_N) = mg +F_Ay         (F_Ay being the y component of the kid's appliedforce)

F_Ay/32.9 = sin(35.1)   

F_Ay = 18.9N

So...    F_N = (17.9)(9.8) +18.9
F_N = 194.3N

b) The frictional force is gonna be the difference between thehorizontal applied force (F_ax) and the horizontal netforce (F_netx)

Use trigonometry again to find horizontal applied force:

F_ax/32.9 = cos(35.1)

F_ax = 26.9N

It gives the horizontal acceleration of the mower, and thatacceleration is due to the net force, not the applied force, so wecan use it to find net force.

F_netx = ma = (17.9)(1.37) = 24.5N

Frictional force is the difference between the two:

F_f = 26.9 - 24.5 = 2.4N

c) This part requires a kinematics equation using the netforce:

V_f = V_i +at               (V_i = 0 since it started from rest)

V_f = at = (1.37)(0.58) =0.79m/s

d)This last one is really tricky, and don't be surprised ifIve made the question more complicated than it was intended to be.(I still think this is right though)

It seems simple: when velocity is constant, net force is zero,so the x component of the kid's applied force equals the frictionalforce. Obviously, the kid's applied force is gonna be less than itoriginally was, because it was making the mower accelerate in thebeginning.

We already calculated frictional force, but the big problemhere is that frictional force depends on normal force. Normal forcedepends on weight of the mower PLUS the y component of the kid'sapplied force. The problem is that the frictional force has changednow that the kid is applying a smaller force, so we don't know whatit is anymore.

The first thing to do is calculate μ, the coefficient offriction, from the equation F_f = μF_N,using the values from the question above. μ will never change- its constant for a certain material (ie. the grass).

2.4 = μ194.3

μ = 0.012

Now, since net force is zero, set up the equationF_f = F_ax    (both these valueswill now be different from the ones in the other questions)

F_f = μF_N, so μF_N =F_ax

Also, F_N = mg + F_ay, so we can now plugthis into one big equation using everything we know:

(0.012)(17.9*9.8 + F_ay) = F_ax

From this now, solve for F_ax in terms ofF_ay:

F_ax = 0.012F_ay + 2.17

There is actually another equation relating the two forcecomponents. Because the mower's handle angle hasnt changed...

F_ay/F_ax = tan(35.1)

From that, F_ax = (F_ay/tan(35.1))

We now have a system of 2 variables and 2 equations, and soits finally solvable:

F_ay/tan(35.1) = 0.012F_ay + 2.17

F_ay = 0.0084F_ay + 1.52

0.992F_ay = 1.52

F_ay = 1.54N

F_ay = Fsin(35.1)

F = 2.67N, the final answer

Hope that wasnt too impossible to follow...

Also sorry I dont have time to check it for mistakes

Answer 2

d) is 2.9N

Fnet = 0

FaxCos35 - Fk = 0

Fax = Fk/Cos35

Fax = 2.9N