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Box slides along horizontal at velocity constant.
FN
Ff
Fp
Fw
on parallel
vc therefore, SF = 0 ; Fp + Ff = 0 ; FF = -Fp
on perpendicular
Rest, therefore, SF = 0 ; FW + FN = 0 ; FN = -Fw
Box slides along ho rizontal at velocity constant.
FN
Ff
Fp
Fw
What is m if the box
is 12 kg and Fp is 34 N?
Fw = mg = 12 kg x -9.8 m/s2 = -120 N,
on perpendicular:
rest, therefore, SF = 0 ;
Fw + FN = 0 ; FN = -Fw
FN = -(-120N) = 120N
on parallel:
vc therefore, SF = 0 ;
Fp + FF = 0 ; Ff = -Fp
FF = -(34N) = -34N
m= FF /FN = -34N/120N
m = 0.28
Box slides along horizontal at velocity constant.
This time the force (Fa) is applied at an upward angle on the box.
Part of Fa acts parallel to the ramp and part acts perpendicular.
The parallel part of Fa pulls the
box forward and the
Fa
FN
perpendicular part pull the
Fv
box upward.
Ff
q
Fp
on parallel:
vc therefore, SF = 0 ;
Fp + FF = 0 ; FF = -Fp
Fw
on perpendicular:
rest, therefore, SF = 0 ;
FW + FN + Fv = 0
FN = -Fw - Fv
Box slides along horizontal at velocity constant.
On parallel:
vc , SF = 0
Fp + FF = 0
FF = -Fp
on perpendicular:
rest, SF = 0
Fw + Fv + FN = 0
FN = -Fw – Fv
FN
Ff
Fa
Fv
q
Fp
What is m if a 65N force applied
at a 300 slides a 12 kg box at
a constant velocity?
Fw
Fv = Sin 300(65N) = +32N
Fp = Cos 300(65N) = +56N
m = FF / FN = FF / (-Fw – Fv) = -56N / -(-120N) – (32N)
m = -56N / 88N = 0.64
Box accelerates
along horizontal
FN
Ff
Fp
Fw
On parallel:
.
accel , . . SF = ma
What forces act in the
direction of the
acceleration?
Fp + FF = ma
Fp = ma - FF
.
..
on perpendicular:
.
rest, . . SF = 0
Fw + FN = 0
FN = – Fw
Box accelerates along horizontal
FN
Ff
Fp = -Ff +
ma
FN = -Fw
Ff = mFN
Fp
Fw
What force is needed to accelerate the 12 kg
box at 2.3 m/s2 if “m” is 0.29?
FN = -Fw = -mg = -(12 kg x -9.8 m/s2 ) = +120 N
Fp = -Ff + ma = -m FN + ma = - -0.29(120N) + 12 kg(2.3 m/s2)
Fp = 62N
Ff is negative because its left,
opposite Fp. FN isn’t negative
But it has to be added to make
Ff negative!
Box accelerates along horizontal.
What acceleration does a 65N
force, applied at a 300 from
the horizontal, give a 12 kg
box if m is 0.29?
Fp + Ff = ma
FN + Fw + Fv = 0
FN = -Fw - Fv
Fa
Fy
Ff
Fp
Fw
FN = -(-120N) - (32N) = 88N
Fp + Ff = ma
Solve for “a”
Fy = Sin 300(65N) = +32N
Fx = Cos 300(65N) = +56N
a = Fp + FF / m
a = Fp + mFN / m
a = Fp + m(Fw -Fv) / m
a = 56N + -(0.29)(88N) / 12kg = 2.5 m/s2
Box slides down the ramp at a constant velocity
on parallel
Ff
FN
Fp
FN’ q
Fw
Fp
FN’ = Cosq Fw
Fp = Sinq Fw
Vc SF = 0
FF + Fp = 0
FF = -Fp
on perpendicular
Rest; SF = 0
FN + FN’ = 0
FN = -FN’
q
Ff
FN’ q
Box slides down the ramp at a constant velocity
Ff = -Fp
FN = -FN’
Fp
FN = Cosq Fw
Fw
Fp = Sinq Fw
Fp
q
Sin q = opp/hyp = 3.9 m/6.5 m
What is m if a 12 kg box slides
down a 6.5 m ramp that is
3.9 m high at a constant velocity?
q = 370
m = Ff /FN = -Fp /FN = - Sin 350 (12kg)-9.8m/s2 /Cos 370 (12kg)-9.8m/s2
m = 0.75
Box accelerates down the ramp
Ff
FN’ q
on parallel
Accel.; SF = ma
Fp + Ff = ma
FN’ = Cosq Fw
Fp
Fp = Sinq Fw
Fw
Ff = m FN
Fp
q
Sin q = opp/hyp = 3.9 m/6.5 m
q = 370
Fp + Ff = ma
a = Fp + Ff /m
What acceleration does a 12 kg
box have if m is 0.45 and the
ramp is 6.5 m long and 3.9 m
high?
Fp = Sin 370 (12kg)-9.8m/s2 = -71N
FN’ = Cos 370 (12kg)-9.8m/s2 = -94N
a = -71N + (42N)/ 12kg = -2.4 m/s2
Ff = 0.45(94N) = 42N
Box accelerates up the ramp
Fa
FN’ q
FN’ = Cosq Fw
Fp
Fw
Ff
Fp = Sinq Fw
Ff = m FN
Fp
q
Sin q = opp/hyp = 3.9 m/6.5 m
q = 370
Fa +Fp + Ff = ma
a = Fa + Fp + Ff /m
on parallel
Accel.; SF = ma
Fa + Fp + Ff = ma
What acceleration does a 12 kg
box have up the ramp if a force of
145 N is applied up the ramp and m is
0.45 and the ramp is 6.5 m long and
3.9 m high?
Fp = Sin 370 (12kg)-9.8m/s2 = -71N
FN’ = Cos 370 (12kg)-9.8m/s2 = -94N
a = 145 N +(-71N) + (-42N)/ 12kg
Ff = 0.45(94N) = -42N
a = 2.7 m/s2
Friction Stopping an Object
Note- there is no forward
force on the person once he
trips and starts sliding!!!
A 55 kg person running
at 6.6 m/s trips and falls
and skids to rest in 2.1 m.
What is m ?
on perpendicular
Ff
FN
Rest, therefore, SF = 0 ; FW + FN = 0
FN = -Fw FN = -(-540 N) = 540 N
The person decelerates once he
Hits the ground because of
Friction. We know v1 = 6.6 m/s,
v2 = 0, and d = 2.1 m, so “vad”
will give us the acceleration.
a = v22 – v12/2d
a = (6.6 m/s)2 – 0/2(2.1m ) = 10 m/s2
on parallel
Fw
accel , SF = ma
-540 N
FF = ma
m FN = ma
m = ma/FN = 55 kg (10 m/s2)/540 m
m = 1.0
Lifting an object
An 11.0 kg box is accelerated
upward at 4.40 m/s2. What
force is needed to do this?
Motion is in the vertical, so
the parallel plane is also
in the vertical.
on parallel
accel , SF = ma
Fw + Fa = ma
Fa = ma – Fw
Fa = (11.0 kg x 4.40 m/s2) – (-108 N) = 156 N
Fa = ?
Fw = -108 N