Clicker Question You are traveling on an elevator up the Sears tower. As you near the - PowerPoint Presentation

Slide1

Clicker Question

You are traveling on an elevator up the Sears tower. As you near the top floor and are slowing down, your accelerationA) is upwardB) is downwardC) is zero

V

mg

N

a

Checkpoint

A box of mass

m is hung by a spring from the ceiling of an elevator. When the elevator is at rest the length of the spring is L = 1 m. If the elevator accelerates upward the length of the spring will be:A) L = 1 mB) L < 1 mC) L > 1 m

m

a

L

“The box is accelerating upward, so the spring force must be greater than gravity.

Therefore, L must be larger than the normal length..”

Accelerating reference frames can confuse…

“There is a pseudo force that pulls the block down, increasing the length of the spring”“The force from the moving elevator will make the box be pushed down, elongating the spring”“accelerating upwards would increase the force downwards and thus increasing the length of the spring”

m

a

L

These arguments may feel good, but

all you really need to remember is this:

acceleration is due to a net force

Mechanics Lecture 6, Slide

4

Clicker ? [continued from Tues]

A cart with mass m2 is connected to a mass m1 using a string that passes over a frictionless pulley, as shown below. The cart is held motionless.

The tension in the string isA) m1gB) m2gC) 0

m

2

m

1

Clicker ? [continued from Tues]

A cart with mass m2 is connected to a mass m1 using a string that passes over a frictionless pulley, as shown below. Initially, the cart is held motionless, but is then released and starts to accelerate.

After the cart is released, the tension in the string isA) = m1gB) > m1gC) < m1g

m

2

m

1

a

a

acceleration is due to a net force

m

1

m

1

g

T

Checkpoint

A block slides on a table pulled by a string attached to a hanging weight. In Case 1 the block slides without friction and in Case 2 there is kinetic friction between the sliding block and the table.

In which case is the tension in the string biggest?A) Case 1 B) Case 2 C) Same

Case 2(With Friction)

Case 1(No Friction)

m

2

m

1

g

m

2

m

1

g

This

is

a difficult question, so let’s step through it

Step 1

In which case is the acceleration of the blocks biggest?A) Case 1 B) Case 2 C) Same

Case 2(With Friction)

Case 1(No Friction)

m

2

m

1

m

2

m

1

In which case is the tension in the string biggest?A) Case 1 B) Case 2 C) Same

Case 2(With Friction)

Case 1(No Friction)

m

2

m

1

m

2

m

1

a

1

a

2

Step 2

Lets work it out

A block (m2) slides on a table pulled by a string attached to a mass (m1) hanging over the side. The coefficient of kinetic friction between the sliding block and the table is mk. What is the tension in the string?

m

2

m

1

Let’s work it out

A block (m2) slides on a table pulled by a string attached to a mass (m1) hanging over the side. The coefficient of kinetic friction between the sliding block and the table is mk. What is the tension in the string?

What is the relationship between the magnitude of the tension of the string at block 2 and the magnitude of the tension in the string at block 1?A) T1 > T2 B) T1 = T2 C) T1 < T2

T

1

T

2

m

2

m

1

g

A helpful aside

:

m

2

m

2g

T

N

m

1

m1g

T

1) FBD

m

2

m

1

f=

m

k

N

ACT

A block (m2) slides on a table pulled by a string attached to a mass (m1) hanging over the side. The coefficient of kinetic friction between the sliding block and the table is mk. What is the tension in the string?

What is the relationship between the magnitudes of the acceleration of the two blocks?A) a1 = a2 B) a1 < a2 C) a1 > a2

m

2

m

1

g

A helpful aside

:

1) FBD2) SF=ma

y: N– m2g=0

x: T – mk N = m2a

y: T– m1g = – m1a

m

2

m

2

g

T

N

f=

mk N

m

1

T

Block 2

Block 1

m

1

g

x

y

m

2

m

1

N

=

m

2

g

T

–

m

k

m

2

g

=

m

2

a

T is smaller when a is bigger

T

= m1g – m1a

a =

m1g – mk m2g

m1 + m2

m

2

m

2

g

T

N

f

m

1

T

1) FBD

2)

S

F

=

ma

m

1

g

x

y

m

2

m

1

T

–

m

1

g

= –

m

1

a

T

– mk m2g = m2a

Solve 2 equations

for 2 unknowns: