Transcript 101KnightChp5

Textbook - Knight
Chapter 5: 33, 42-45
Chap 5:33 pg 156
What is the minimum
downward force on the
box which will keep if
from slipping. The
coefficients of static and
kinetic friction between
the box and the floor are
.35 and .25 respectively.
• Not slipping means
static: Ffrict = μsFnorm
• Fnorm = F + mg = F+300
• Ffrict = 125
• Ffrict = μsFnorm
• 125 = .35(F+300)
• Solve
Chap 5:42 pg 157
• What is the tension on
the rope ?
• Forces on 60 kg
– Gravity = mg = 600N down
– Tension of Rope = T
– Fnet = T − 600 = ma = 0
• T = 600 newtons
• Forces on 100 kg
–
–
–
–
Gravity = mg = 1000N down
Tension of Rope = T = 600N up
Foor pushing up = Ffloor
Fnet = T − 1000 + Ffloor = ma = 0
• Ffloor = 1000 − T
•
= 1000 − 600 = 400
Chap 5:43 pg 157
A 2-meter-long, 500 g
rope pulls a 10kg block
of ice across a
horizontal, frictionless
surface. The block
accelerates at 2 m/s2.
How much force pulls on
(a) the block
(b) the rope
•
•
•
•
Frictionless
Rope pulls horizontally
Normal (vertical) forces cancel
Forces on the block (10kg)
– left end of rope (pulls to right)
– vertical forces cancel
–
ma = Fnet = left end of rope
–
= 10*2 = 20 newtons right
• Forces on rope (m = .5 kg)
– left end = 20 newtons left
– right end = R
– Fnet = −20 + R = ma
–
−20 + R = .5*2 = 1 newton
–
−20 + R = 1
–
R = 21 newtons
Chap 5:44 pg 157
• Two 1-kg blocks are connected by a rope. A
second rope hangs beneath the lower block.
Both ropes have a mass of 250g. The entire
assembly accelerates up at 3 m/s2
• All forces are vertical
• Start with Rope 2 (m = .25 kg)
– Force at bottom =
– Force at top = T =
• Block B
–
–
–
–
(m = 1 kg)
Rope 2 pulls down =
Gravity = mg =
Rope 1 pulls up
Fnet = ma =
down