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F=ma. Newton's 2nd law.
Thursday, 10 March 2011
1:33 PM
Newton's Laws of Motion
1. If no NET force on a body, then it travels in straight line
at constant velocity (including zero velocity). No accel.
2. F=ma
3. Every Action has equal and opposite reaction.
If NET force, (or total of all forces) is zero, then body is in equilibrium.
Fw
F air
Fa
EXAMPLE
100km/h, level, straight.
Sum of forces = 0 WHY?
F = ma = 0
F = ma
F = ma
Fb
F road
This is better, because Newton's 2nd law
means the NET force = ma.
F = 1090 * 1.4 = 1526 N
(kg) (m/s2) (N)
F=ma3 Page 1
F = ma
F = ma
F = (2077*1000) * (2.7 * 9.81)
= 5.5013E7 (same as 5.5013 * 10^7)
(on calculator use EXP button)
Now use ENG button, to convert to nearest
power of 3,6,9, etc.
This matches SI units ok k, M,G etc.
Answer = 55.013E6 N (55 MN)
a=F/m
= 2600 / 1090 = 2.3853 m/s2
Now get velocity; (Linear Motion)
F=ma3 Page 2
So a = 2.3853 m/s2 , v0 = 0, t = 15, v1 = ?
Use eqn (4): v1 = v0 + a*t
= 0 + 15 * 2.3853 = 35.7795 m/s
Convert to km/h = m/s x 3.6
35.7795 * 3.6 = 128.8062 km/h
1 km/h = 1000 m/h = 1000/3600 m/s
1 m/s = 0.001 km/s = 0.001/1/3600 km/h
= 3.6 km/h
F=ma3 Page 3
F=ma with Gravity
Thursday, 10 March 2011
2:26 PM
Man in lift...
A person of mass 65kg stands on a force scale
inside a lift. What is the reading on the scale (N)
when;
(a) Lift is stationary
F=ma... Does man have a net force?
Net Force = zero (equal and opposite)
Scale reads = 65*9.81 = 637.65 N
F = ma = 65*0 = 0 N Inertial force
Total = 637.65 + 0 = 637.65 N
(b) Lift accelerates upward at 2 m/s2
F=ma... Does man have a net force?
F floor > F weight
F = ma not zero!
F = ma = 65*2 = 130 N inertial force
Total at scale = 637.65 + 130 = 767.65 N
F = ma
Ffloor - Fw = ma
F floor = ma + Fw
= 130 + 637.65 = 767.65 N
(c) Lift is moving upward at 4.5 m/s (accel = 0)
Inertial force = ma = 65 *0 = 0 N
F = ma
Ffloor - Fw = ma
F floor = ma + Fw
= 0 + 637.65 = 637.65 N
(d) Lift decelerates to the top floor at -2 m/s2
Inertial force = ma = 65 *-2 = -130 N
F=ma3 Page 4
(d) Lift decelerates to the top floor at -2 m/s2
Inertial force = ma = 65 *-2 = -130 N
F = ma
Ffloor - Fw = ma
F floor = ma + Fw
= -130 + 637.65 = 507.65 N
(feels lighter!)
(e) Lift accelerates downward at 2 m/s2
Inertial force = ma = 65 *-2 = -130 N
F = ma
Ffloor - Fw = ma
F floor = ma + Fw
= -130 + 637.65 = 507.65 N
(feels lighter!)
(f) Lift is moving downward at 4.5 m/s (accel = 0)
Inertial force = ma = 65 *0 = 0 N
F = ma
Ffloor - Fw = ma
F floor = ma + Fw
= 0 + 637.65 = 637.65 N
(g) Lift decelerates to the bottom floor at -2 m/s2
Downwards deceleration = downwards negative
acceleration = upwards acceleration..
F = ma
Ffloor - Fw = ma
F floor = ma + Fw
= 130 + 637.65 = 767.65 N
F=ma3 Page 5
F floor
Fw
F = ma
Fth - Fw = ma
Fth = ma + Fw
So
Inertial Force;
ma = (2100*1000) *(2.8*9.81) = 57682800 N
Weight Force;
Fw = (2100*1000) * 9.81 = 20601000 N
Thrust Force;
Fth = ma + Fw
= 57682800 + 20601000 = 78283800 N
F=ma3 Page 6
Fth
F=ma and Friction
Thursday, 10 March 2011
3:01 PM
Example
Coeff = Ff / Fn
Ff = 145 N
Fn = 64 * 9.81 = 627.84 N
Coeff = Ff / Fn = 145 / 627.84 = 0.231
F = ma
Total F = ma
Ff = 0.33 * (3.3 *9.81) = 10.6831 N
Summing in horizontal only...
F - Ff = ma
F = ma + Ff
= (3.3 * 3.1) + 10.6831
F=ma3 Page 7
= (3.3 * 3.1) + 10.6831
= 20.9131 N
F=ma3 Page 8
INCLINED PLANE
Tuesday, 22 February 2011
5:10 PM
Any question relating to inclined planes
or friction...
STEP 1: Project all forces onto parallel
and perpendicular to the plane
ANGLE
Convert weight;
Wn = W cos(theta)
Wp= W sin( theta)
------------------------------------------------Example: 58kg trolley, 1 in 15 gradient
Find angle:
Angle = atan (1/15) = 3.814 o
Find weight:
W = 58 * 9.81 = 568.98 N
Wp= 568.98*sin(3.814) = 37.847 N
Now get acceleration;
F = ma, so a = F/m
a = 37.847 / 58 = 0.6525 m/s2
---------------------------------------------------
Wn = W cos(theta)
W
Wp= W sin( theta)
Acceleration with Friction:
Block = 8.4 kg, coeff = 0.3, accel = 5.2 m/s 2
Find P
W = 8.4 * 9.81 = 82.404 N
Fn = W = 82.404 N
F f = coeff * Fn = 0.3 * 82.404 = 24.7212 N
Friction force always opposes direction of MOTION.
Net (total) force F = ma = 8.4 * 5.2 = 43.68 N
(Total) F = P - Ff
P - Ff = 43.68 N
So P = 43.68 + Ff = 43.68 + 24.7212 = 68.4012 N
---------------------------------------------------
F=ma3 Page 9
W
P
Ff
Fn
MULTIPLE BODIES!
Tuesday, 22 February 2011
5:42 PM
Masses = 18.5 kg, coeff = 0.4, P = 98N
P
Start off by combined body, to find a;
m= 18.5 * 2 = 37 kg
W = 9.81 *37 = 362.97 N
Get Ff;
Ff = 0.4 * 362.97 = 145.188 N
Total F = P - Ff = 98 - 145.188
OOPS - it doesn't move!
- ------------------------------Now take front block only!
m = 18.5 kg
W = 9.81 * 18.5 = 181.485 N
Ff = 0.4 *181.485 = 72.594 N
W
P
FL
Ff
Fn
Do horizontal forces;
Ff + FL - P = 0 so FL = P - Ff
FL = 98 - 72.594 = 25.406 N
---------------------------------Coeff = 0.07, Ma =13kg, Mb = 6kg, P = 72N
Find accel.
W = 9.81*(13+6) = 186.39 N which = Fn
Ff = coeff *Fn = 0.07 * 186.39 = 13.0473 N
Total F = ma
P - Ff = ma so a = (P-Ff)/M
a = (72 - 13.0473) / (13 + 6 ) = 3.1028 m/s2
Ma = 15kg, Mb = 7kg, coeff = 0.3, P=76N
Find force exerted onto block B.
Do FBD of both blocks as one to get a;
W = mg = (15 + 7) *9.81 = 215.82 N
Ff = coeff * Fn = 0.3 * 215.82 = 64.746 N
Total F = P - Ff = 76 - 64.746 = 11.254 N
a = F/m = 11.254 / (15 + 7 ) = 0.5115 m/s2
Now do FBD for block B to get Force onto B;
F=ma3 Page 10
P
Ff
P
Ff
Find Ff:
Total F = ma = 7 * 0.5115 = 3.5805 N
Find Ff;
Ff = coeff * Fn = 0.3 * (9.81*7) = 20.601 N
Total F = Fab - Ff
Fab = F + Ff = 3.5805 + 20.601 = 24.1815 N
W
Fab
Ff
Fn
FBD for Block B to get "?"
F=ma3 Page 11
WORSEST: BIG BAD AND UGLY QUESTION
Tuesday, 22 February 2011
6:42 PM
Pulley Rule: Tension is the same on both sides!
Also: We know acceleration is same for both masses.
Ma = 1.5kg, Mb = 3.7 kg
Wa = 9.81 * 1.5 = 14.715 N
Wb = 9.81 * 3.7 = 36.297 N
Start with FBD of mass A.
Total F = ma
But Total F = Ft - Wa
So Ft - Wa = 1.5 * a
Ft - 14.715 = 1.5* a -----------------(1)
Now FBD for mass B; F = ma
Wb - Ft = 3.7 * a
36.297 - Ft = 3.7 * a -------------------(2)
Solve by simultaneous equations.
From (1) ; Ft = 1.5a +14.715
Subs into (2)...
36.297 - (1.5a +14.715) = 3.7 * a
36.297 - 1.5a -14.715 = 3.7a
36.297 -14.715 = 3.7a + 1.5a
21.582 = 5.2 a
a = 21.582/5.2 = 4.1504 m/s2
Now subs back into (1) to get tension...
Ft = 1.5a +14.715
Ft = 1.5*4.1504 +14.715 = 20.9406 N
F=ma3 Page 12
Ft
Ft
Wa
Wb
FRICTION AND INCLINED PLANE
Tuesday, 8 March 2011
5:36 PM
p151
where
is the friction force.
is the coefficient of friction. (depends on materials)
is the normal force exerted between the surfaces, (which
is equal to W in this diagram).
Always take components normal and parallel to plane.
Q11.1:
150kg box, 15 degs, coeff=
0.35, What force to push
up hill?
Fw = 9.81*150 = 1471.5 N
Fwx = 1471.5* sin(15) = 380.85 N
(Gravity force downhill!)
Fwy = 1471.5*cos(15) = 1421.36 N
(Same as Fn: Use this to find friction)
Ff = 0.35 * 1421.36 = 497.48 N
(Always oppose friction!)
The following questions are CONSTANT VELOCITY (No accel)
Total force to raise block = Gravity force + friction force
F=ma3 Page 13
Total force to raise block = Gravity force + friction force
= 380.85 + 497.48 = 878.33 N
What about DOWN the hill?
Total force to lower block = Friction force - Gravity force
= 497.48 - 380.85 = 116.63 N
What if there was NO friction?
Force to raise block = Fwx = 380.85 N
Force to lower block = Fwx = 380.85 N
11.3 m=20kg, 25 degs is angle of repose. What force required to
hold it from slipping at 35 degs?
Coeff = Ff/Fn = F sin(ang) / F cos(ang) = sin/cos = tan
Sin = a/c
Cos= b/c
Sin/Cos = a/c / b/c = a/c * c/b = a/b = Tan. See!
So Coeff of friction = Tan(angle of repose)
If Ang = 25, then coeff = tan(25) = 0.4663
Now set angle at 35 degs...
Fw = 9.81 * 20 = 196.2 N
Fwx = 196.2 * sin(35) = 112.54 N (Fwp)
Fwy = 196.2 * cos(35) = 160.72 N (Fwn)
Ff = 0.4663 * 160.72 = 74.944 N
Force to hold it still (same as force for downhill)
Total force to lower block = Friction force - Gravity force
F=ma3 Page 14
Total force to lower block = Friction force - Gravity force
= 74.944 - 112.54 = -37.596 N (downhill)
= 37.596 N (uphill)
Q11.4: What force required if horizontal?
Unknown force F1
This adds weight to the normal
component, which increases the
friction, which lowers the force to
hold it still!
Total Fy = F1y+ Fwy
Sin(35) = F1y/F1 so F1y = F1*sin(35)
Sin(35) = 0.57358
So F1y = 0.57358 F1 .................(a)
Cos(35) = 0.81915
F1x = 0.81915 F1 .................(b)
Get Fn to find Ff;
Fn = Fwy + F1y = 160.72 + 0.57358 F1
Ff = coeff* Fn = 0.4663 * (160.72 + 0.57358 F1)
Force to hold it still (same as force for downhill)
Total force to lower block = Friction force - Gravity force
= 0.4663 * (160.72 + 0.57358 F1) - 112.54
(Help! We can't get force F1 until we know friction Ff, but force
changes the normal force Fn which then changes friction Ff... Drat! )
(So once we get forces on an ANGLE to the plane, things get
complicated. Check out the next section using the graphical method
for solving friction problems...)
F=ma3 Page 15
FRICTION BY GRAPHICAL METHOD - (Very easy method)
Tuesday, 8 March 2011
6:47 PM
Angle of friction...(p155)
A of F = atan(coeff)
m=20kg, 25 degs is angle of repose. What force required to hold
it from slipping at 35 degs?
A of F = atan(0.4663) = 25 degs ! (Same as angle of repose)
Since we are just stopping it from going DOWNHILL, we use the
downhill friction. (which is pointing 25 degs up from the normal)
FBD
Fr = surface reaction force
Fr @ 65+35 = 100
F1? @0
FBD
Fw @270
F=ma3 Page 16
FBD
Fw @270
Force
Polygon
F=ma3 Page 17