box slides along horizontal at velocity constant. ffff fpfp fwfw v c therefore, f = 0 ; f p + f f =...
TRANSCRIPT
Box slides along horizontal at velocity constant.
Ff Fp
Fw
vc therefore, F = 0 ; Fp + Ff = 0 ; FF = -Fp
Rest, therefore, F = 0 ; FW + FN = 0 ; FN = -Fw
FN
on parallel
on perpendicular
Box slides along ho rizontal at velocity constant.
Ff Fp
Fw on parallel:vc therefore, F = 0 ; Fp + FF = 0 ; Ff = -Fp
FF = -(34N) = -34N
What is if the boxis 12 kg and Fp is 34 N?
Fw = mg = 12 kg x -9.8 m/s2 = -120 N,
= FF /FN = -34N/120N = 0.28
FN
on perpendicular:rest, therefore, F = 0 ; Fw + FN = 0 ; FN = -Fw
FN = -(-120N) = 120N
Box slides along horizontal at velocity constant.
Ff
Fw
vc therefore, F = 0 ; Fp + FF = 0 ; FF = -Fp
Fa
Fp
Fv
FN
on parallel:on perpendicular:
rest, therefore, F = 0 ; FW + FN + Fv = 0 FN = -Fw - Fv
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 perpendicular part pull thebox upward.
Box slides along horizontal at velocity constant.
Ff
Fp
Fw
On parallel: vc , F = 0 Fp + FF = 0FF = -Fp
Fa
Fv
What is if a 65N force appliedat a 300 slides a 12 kg box ata constant velocity?
Fv = Sin 300(65N) = +32N
Fp = Cos 300(65N) = +56N
= FF / FN = FF / (-Fw – Fv) = -56N / -(-120N) – (32N)
FN
on perpendicular: rest, F = 0 Fw + Fv + FN = 0FN = -Fw – Fv
Box accelerates along horizontal
Ff Fp
Fw
FN
On parallel: accel , F = maWhat forces act in the direction of theacceleration? Fp + FF = ma Fp = ma - FF
on perpendicular: rest, F = 0 Fw + FN = 0FN = – Fw
.. .
.. .
.. .
Box accelerates along horizontal
Ff Fp
Fw
Fp = -Ff + ma
FN = -Fw
Ff = FN
What force is needed to accelerate the 12 kgbox at 2.3 m/s2 if “”is 0.29?
FN = -Fw = -mg = -(12 kg x -9.8 m/s2 ) = +120 N
Fp = 62N
FN
Fp = -Ff + ma = - FN + ma = - -0.29(120N) + 12 kg(2.3 m/s2)Ff is negative because its left, opposite Fp. FN isn’t negativeBut it has to be added to make Ff negative!
Box accelerates along horizontal.
Ff Fp
Fw
Fp + Ff = maFN + Fw + Fv = 0
Fa
Fy
What acceleration does a 65N force, applied at a 300 from the horizontal, give a 12 kg box if is 0.29?
Fy = Sin 300(65N) = +32N
Fx = Cos 300(65N) = +56N
Fp + Ff = maSolve for “a”
a = Fp + FF / m
FN = -(-120N) - (32N) = 88N
a = Fp + FN / ma = Fp + Fw -Fv) / m
a = 56N + -(0.29)(88N) / 12kg = 2.5 m/s2
FN = -Fw - Fv
Box slides down the ramp at a constant velocity
Fw
Ff
Fp
FF = -Fp
FN’ = Cos Fw
Fp
FN’
Fp = Sin Fw
FN
FN = -FN’
on parallel
on perpendicular
Vc F = 0
FF + Fp = 0
Rest; F = 0
FN + FN’ = 0
Box slides down the ramp at a constant velocity
Ff = -Fp
FN = Cos Fw
Fw
Ff
Fp
Fp
FN’ Fp = Sin Fw
What is if a 12 kg box slidesdown a 6.5 m ramp that is 3.9 m high at a constant velocity?
Sin opp/hyp = 3.9 m/6.5 m
= 0.75
FN = -FN’
Ff /FN = -Fp /FN = - Sin 350 (12kg)-9.8m/s2 /Cos 370 (12kg)-9.8m/s2
Box accelerates down the ramp
Fp + Ff = ma
FN’ = Cos Fw
Fw
Ff
Fp
Fp
FN’ Fp = Sin Fw
What acceleration does a 12 kgbox have if is 0.45 and the ramp is 6.5 m long and 3.9 m high?
Sin opp/hyp = 3.9 m/6.5 m
Fp + Ff = ma
a = Fp + Ff /m
Fp = Sin 370 (12kg)-9.8m/s2 = -71N
FN’ = Cos 370 (12kg)-9.8m/s2 = -94N
Ff = FN
Ff = 0.45(94N) = 42Na = -71N + (42N)/ 12kg = -2.4 m/s2
on parallelAccel.; F = ma
Box accelerates up the ramp
Fa + Fp + Ff = ma
FN’ = Cos Fw
FwFf
Fp
Fp
FN’ Fp = Sin Fw
What acceleration does a 12 kgbox have up the ramp if a force of 145 N is applied up the ramp and is 0.45 and the ramp is 6.5 m long and 3.9 m high?
Sin opp/hyp = 3.9 m/6.5 m
Fa +Fp + Ff = ma
a = Fa + Fp + Ff /m
Fp = Sin 370 (12kg)-9.8m/s2 = -71N
FN’ = Cos 370 (12kg)-9.8m/s2 = -94N
Ff = FN
Ff = 0.45(94N) = -42Na = 145 N +(-71N) + (-42N)/ 12kg
on parallelAccel.; F = maFa
a = 2.7 m/s2
Ff
Fw
FN
Friction Stopping an Object
A 55 kg person runningat 6.6 m/s trips and fallsand skids to rest in 2.1 m.What is ?
Note- there is no forwardforce on the person once hetrips and starts sliding!!!
on perpendicular
on parallel
Rest, therefore, F = 0 ; FW + FN = 0FN = -Fw FN = -(-540 N) = 540 N
accel , F = ma FF = ma FN = ma = ma/FN = 55 kg (10 m/s2)/540 m = 1.0
-540 N
The person decelerates once heHits the ground because ofFriction. We know v1 = 6.6 m/s,v2 = 0, and d = 2.1 m, so “vad”will give us the acceleration.a = v2
2 – v12/2d
a = (6.6 m/s)2 – 0/2(2.1m ) = 10 m/s2
Motion is in the vertical, sothe parallel plane is alsoin the vertical.
Lifting an object
An 11.0 kg box is acceleratedupward at 4.40 m/s2. Whatforce is needed to do this?
Fw = -108 N
Fa = ?
Not
e, th
ere a
re n
o fo
rces
On
the h
oriz
onta
l!!!
on parallel
accel , F = maFw + Fa = maFa = ma – Fw
Fa = (11.0 kg x 4.40 m/s2) – (-108 N) = 156 N