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