Review for Dynamics testPage 1 - Net force with weightPage 2 - Friction on a level surfacePage 3 - Inclined PlanePage 4 - Statics

Page 1 - elevator problem

Net Force – Example 3 Using Weight

TOC

5.0 kg

35 N

Find the acceleration(on Earth)

Net Force – Example 3 Using Weight

Draw a Free Body Diagram:

TOC

5.0 kg

35 N

Don’t Forget the weight:F = ma = 5.0*9.8 = 49 N

-49 N

Net Force – Example 3 Using Weight

F = ma35 N – 49 N = (5.0 kg)a-14 N = (5.0 kg)a a = -2.8 m/s/s

TOC

5.0 kg

35 N

-49 N

Whiteboards: Using Weight1 | 2 | 3 | 4 | 5

TOC

2.7 m/s/s W

8.0 kg

100. N

F = ma, weight = (8.0 kg)(9.80 N/kg)= 78.4 N downMaking up +<100. N - 78.4> = (8.0kg)a21.6 N = (8.0kg)aa = 2.7 m/s/s

Find the acceleration:

-1.8 m/s/s W

15.0 kg

120. N

F = ma, wt = (15.0 kg)(9.8 N/kg) = 147 N down<120. N - 147 N> = (15.0kg)a-27 N = (15.0kg)aa = -1.8 m/s/s It accelerates down

Find the acceleration:

180 N W

16 kg

F

F = ma, wt = (16 kg)(9.8 N/kg) = 156.8 N down<F – 156.8 N> = (16.0 kg)(+1.5 m/s/s)F – 156.8 N = 24 NF = 180.8 N = 180 N

Find the force:

a = 1.5 m/s/s(upward)

636 N W

120. kg

F

F = ma, wt = 1176 N downward<F – 1176 N> = (120. kg)(-4.50 m/s/s)F – 1176 N = -540 NF = 636 N

Find the force:

a = -4.50 m/s/s(DOWNWARD)

16,900 N W

120. kg

FFirst, suvat:s = -1.85 m, u = -22.0 m/s, v = 0, a = ?use v2 = u2 + 2as, a = +130.81 m/s/s

F = ma, wt = 1176 N downward<F – 1176 N> = (120. kg)(+130.81 m/s/s)F – 1176 N = 15697 NF = 16873.2973 = 16,900 N

This box is going downwards at 22.0 m/s and is stopped in a distance of 1.85 m. What must be the upwards force acting on it to stop it?

Page 2 - Friction on the level

FFr = FNForce of Friction in N

Coefficient of Friction. 0 < < 1(Specific to a surface) - in your book (Table 4-2)

Normal Force - Force exerted by a surface to maintain its integrity

Usually the weight (level surfaces)

Kinetic Friction - Force needed to keep it going at a constant velocity.

FFr = kFN

Always in opposition to velocity (Demo, example calculation)

Static Friction - Force needed to start motion.FFr < sFN

Keeps the object from moving if it can.Only relevant when object is stationary.Always in opposition to applied force.Calculated value is a maximum(Demo, example calculation, examples of less than maximum)

Whiteboards: Friction

1 | 2 | 3 | 4 | 5 | 6

TOC

What is the force needed to drag a 12 kg chunk of rubber at a constant velocity across dry concrete?F = ma, FFr = kFN

FN = weight = mg = (12 kg)(9.8 N/kg) = 117.6 NFFr = kFN = (.8)(117.6 N) = 94.08 N = 90 N

90 N W

What is the force needed to start a 150 kg cart sliding across wet concrete from rest if the wheels are locked up?F = ma, FFr < sFN

FN = weight = mg = (150 kg)(9.8 N/kg) = 1470 NFFr = kFN = (.7)(1470) = 1029 N = 1000 N

1000 N W

s = .62, k = .48What is the acceleration if there is a force of 72 N in the direction an 8.5 kg block is already sliding?

FFr = kFN, FN = mg, FFr = kmgFFr = (.48)(8.5 kg)(9.8 N/kg) = 39.984 NF = ma<72 N - 39.984 N> = (8.5 kg)a, a = 3.77 = 3.8 ms-2

3.8 m/s/s W

72 N8.5 kg

v

FFr

s = .62, k = .48A 22 kg block is sliding on a level surface initially at 12 m/s. What time to stop?

FFr = kFN, FN = mg, FFr = kmgFFr = (.48)(22 kg)(9.8 N/kg) = 103.488 NF = ma< -103.488 N> = (22 kg)a, a = -4.704 ms-2

v = u + at, v = 0, u = 12, a = -4.704 ms-2, t = 2.55 s = 2.6 s

2.6 s W

22 kg

v=12m/s

FFr

s = .62, k = .48

A 6.5 kg box accelerates and moves to the right at 3.2 m/s/s, what force must be applied?

FFr = kFN, m = 6.5 kgFFr = (.48)(6.5 kg)(9.8 N/kg) = 30.576 NF = ma< F - 30.576 N > = (6.5 kg)(3.2 ms-2), F = 51.376 = 51 N

51 N W

F = ?FFr6.5 kg

v a = 3.2 ms-2

s = .62, k = .48A 22 kg block is sliding on a level surface initially at 12 m/s stops in 2.1 seconds. What external force is acting on it besides friction??

v = u + at, v = 0, u = 12 m/s, t = 2.1 s, a = -5.7143 ms-2

FFr = kFN, FN = mg, FFr = kmgFFr = (.48)(22 kg)(9.8 N/kg) = 103.488 NF = ma< -103.488 N + F> = (22 kg)(-5.7143 ms-2), F = -22 N (left)

-22 N (to the left) W

22 kg

v=12m/s

FFrF = ?

s = .62, k = .48

A force of 35 N in the direction of motion accelerates a block at 1.2 m/s/s in the same direction What is the mass of the block?

FFr = kFN, FN = mg, FFr = kmgFFr = (.48)m(9.8 ms-2) = m(4.704 ms-2)F = ma< 35 N - m(4.704 ms-2) > = m(1.2 ms-2)35 N = m(4.704 ms-2) + m(1.2 ms-2) = m(4.704 ms-2 + 1.2 ms-2)m = (35 N)/(5.904 ms-2) = 5.928 kg = 5.9 kg

5.9 kg W

35 NFFrm

v a = 1.2 ms-2

Page 3 - Inclined planes

mg

Fperp = mgcos

F|| = mgsin

FN = mgcos(Causes friction)

(FFr = kFN )

And the plane pushes back(It doesn’t break)

Since we know the angle, we can calculate the components

(Acts down the plane)

Whiteboards: Inclines with friction

1 | 2 | 3 | 4 | 5

TOC

Find Fperp, F||, the kinetic and maximum static friction:

F|| = mgsin() = (3.52 kg)(9.8 N/kg)sin(42.0o) = 23.08 N

Fperp = mgcos() = (3.52 kg)(9.8 N/kg)cos(42.0o) = 25.64 NFFr(kinetic) = kFN = (.37)(25.64 N) = 9.49 NFFr(static) < sFN = (.82)(25.64 N) = 21.02 N

26 N, 23 N, 9.5 N, 21 N W

= 42.0o

3.52 kg

s = .82k = .37

•No, it will not stay. The maximum static (FFr(static)<21.02 N) friction is smaller than the gravity parallel to the plane (F|| =23.08 N)

No, Blue W

= 42.0o

3.52 kg

s = .82k = .37

F|| = 23.08 N, FFr(kinetic) = 9.49 N, FFr(static) < 21.02 N

•Will it stay on the plane if u = 0?

+

-

•Down the plane is negative, so we have the parallel force down (-) the plane, and kinetic friction up (+) the plane:

-3.9 m/s/s W

= 42.0o

3.52 kg

s = .82k = .37

F|| = 23.08 N, FFr(kinetic) = 9.49 N, FFr(static) < 21.02 N

•What will be its acceleration down the plane if it is sliding down the plane?

+

-

•<-23.08 N + 9.49 N> = (3.52 kg)a, a = -3.86 m/s/s = -3.9 m/s/s

-23.08 N

+9.49 N

<-23.08 N - 9.49 N> = (3.52 kg)a, a = -9.25 m/s/s = -9.3 m/s/sv = 0, u = 5.0 m/s, a = -9.25 m/s/s, v2=u2+2as, s = 1.35 m = 1.4 m

-9.3 m/s/s, 1.4 m W

= 42.0o

3.52 kg

s = .82k = .37

F|| = 23.08 N, FFr(kinetic) = 9.49 N, FFr(static) < 21.02 N

The block is given an initial velocity of 5.0 m/s up the plane. What is its acceleration as it slides up the plane? How far up does it go?

-

+

-23.08 N

-9.49 N

As it slides up the plane, the parallel force is down (-) the plane (always), and since the velocity is up the plane, the kinetic friction is now also down (-) the plane:

So now we have an unknown force F probably up the plane, and the parallel force as always down (-) the plane, and friction which is down (-) the plane, because the velocity is up the plane, as well as an acceleration that is up (+) the plane:

56 N W

= 42.0o

3.52 kg

s = .82k = .37

F|| = 23.08 N, FFr(kinetic) = 9.49 N, FFr(static) < 21.02 N

What force in what direction will make it accelerate and slide up the plane at 6.7 m/s/s?

-

+

-23.08 N

-9.49 N

F = ?

<-23.08 N - 9.49 N + F> = (3.52 kg)(+6.7 m/s/s), F = 56.15 N = 56 N

4.8 N W

= 42.0o

3.52 kg

s = .82k = .37

F|| = 23.08 N, FFr(kinetic) = 9.49 N, FFr(static) < 21.02 N

What force in what direction will make it accelerate and slide down the plane at 2.5 m/s/s?

+

-

So now we have an unknown force F , and the parallel force as always down (-) the plane, and friction which is up (+) the plane, because the velocity is down the plane, as well as an acceleration that is down (-) the plane:

a = -2.5 m/s/s

<-23.08 N + 9.49 N + F> = (3.52 kg)(-2.5 m/s/s), F = 4.797 N = +4.8 N

-23.08 N

+9.49 N

F = ?

Page 4 - Statics

Force equilibrium:

Step By Step:1. Draw Picture2. Calculate weights3. Express/calculate components4. Set up a <sum of all forces> = 0 equation

for x and another for the y direction5. Do math.

54 kg

If F1 is 185 N, what is F2 ?

F1

F2

529.2 N

First - the box weighs (54 kg)(9.8 N/kg) = 529.2N

This is a downward force

Next, since there are no forces in the x direction, and there are no components to make, let’s set up our Y equation:F1 + F2 - 529.2 = 0, but since F1 = 185 N185 N + F2 - 529.2 = 0, so F2 = 344.2 N

(Two questions like this)

How to set up torque equilibrium:1. Pick a point to torque about.2. Express all torques:3. +rF +rF+rF… = 0

• + is CW, - is ACW• r is distance from pivot

4. Do math

5.25 NF = ?

2.15 m 5.82 m

1. Torque about the pivot point2 and 3. (2.15 m)(5.25 N) - (5.82 m)F = 0F = 1.94 N (mech. adv.)

WhiteboardsSimple Torque Equilibrium

1 | 2 | 3

(One question like this)

315 N 87.5 N

12 m r = ?

Find the missing distance. Torque about the pivot point.

43 m

(315 N)(12 m) - (87.5 N)r = 0r = 43.2 m = 43 m

W

34 N

F = ?

1.5 m 6.7 m

Find the missing force. Torque about the pivot point. (Be careful of the way the distances are marked)

6.2 N

(34 N)(1.5 m) - (8.2 N)F = 0F = 6.2 N

W

512 N

F = ?

2.0 m 4.5 m

Find the missing Force. Torque about the pivot point.

360 N

-(512 N)(2.0 m) - (481 N)(5.1 m) + F(9.6 m) = 0F = 362 N = 360 N

W

481 N

3.1 m

Force Equilibrium:1. Draw picture2. Calculate weights3. Draw arrows for forces.

(weights of beams act at their center of gravity)

4. Make components5. Set up sum Fx = 0, sum Fy = 0

Torque Equilibrium:1. Pick a Pivot Point

(at location of unknown force)2. Express all torques:3. +rF +rF+rF… = 0

+ is CW, - is ACWr is distance from pivot

Do Math

To do a combined force and torque problem:

Whiteboards: Torque and force

2a | 2b | 2c

TOC(Pretty much kinda exactly this problem)

T1 + T2 -509.6 N - 754.6 N = 0

Step 1 - Set up your vertical force equation

52 kg

77 kgT1 T2

Beam is 18.0 m long, person is 5.0 m from the right side. Find the two tensions in the cables at either end.

T1 and T2 are up, and the beam and person weights are down:Beam: -(52 kg)(9.8 N/kg) = -509.6 N (down)Person: -(77 kg)(9.8 N/kg) = -754.6 N (down)T1 + T2 -509.6 N - 754.6 N = 0

+4586.4 Nm + 9809.8 Nm - (18.0 m)(T2) = 0

Step 2 - Let’s torque about the left sideSet up your torque equation: torque = rF

52 kg

77 kgT1 T2

Beam is 18.0 m long, person is 5.0 m from the right side. Find the two tensions in the cables at either end.

T1 = 0 Nm torque, (r = 0)Beam: (9.0 m)(509.6 N) = +4586.4 Nm (CW)Person: (13.0 m)(754.6 N) = +9809.8 Nm (CW)T2: T2 at 18.0 m = -(18.0 m)(T2) (ACW)Finally:+4586.4 Nm + 9809.8 Nm - (18.0 m)(T2) = 0

9.0 m 13.0 m 18.0 m

509.6 N 754.6 N

T2 = 799.8 N T1 =464.4 N

Step 3 - Math time. Solve these equations for T1

and T2:+4586.4 Nm + 9809.8 Nm - (18.0 m)(T2) = 0T1 + T2 -509.6 N - 754.6 N = 0

52 kg

77 kgT1 T2

Beam is 18.0 m long, person is 5.0 m from the right side. Find the two tensions in the cables at either end.

+4586.4 Nm + 9809.8 Nm = (18.0 m)(T2), T2 = 799.8 NT1 + 799.8 N -509.6 N - 754.6 N = 0, T1 = 464.4 N