Review for Dynamics test Page 1Page 1 - Net force with weight Page 2Page 2 - Friction on a level surface Page 3 Page 3 - Inclined Plane Page 4 Page 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 = ma 35 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 Weight 11 | 2 | 3 | 4 | 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 down Making up + = (8.0kg)a 21.6 N = (8.0kg)a a = 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 = (15.0kg)a -27 N = (15.0kg)a a = -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) = N down = (16.0 kg)(+1.5 m/s/s) F – N = 24 N F = 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 = (120. kg)(-4.50 m/s/s) F – 1176 N = -540 N F = 636 N Find the force: a = m/s/s (DOWNWARD)

16,900 N W 120. kg F First, suvat: s = m, u = m/s, v = 0, a = ? use v 2 = u 2 + 2as, a = m/s/s F = ma, wt = 1176 N downward = (120. kg)( m/s/s) F – 1176 N = N F = = 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

F Fr =  F N Force 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)

K inetic Friction - Force needed to k eep it going at a constant velocity. F Fr =  k F N Always in opposition to velocity (Demo, example calculation) St atic Friction - Force needed to st art motion. F Fr <  s F N 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 11 | 2 | 3 | 4 | 5 | TOC

What is the force needed to drag a 12 kg chunk of rubber at a constant velocity across dry concrete? F = ma, F Fr =  k F N F N = weight = mg = (12 kg)(9.8 N/kg) = N F Fr =  k F N = (.8)(117.6 N) = 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, F Fr <  s F N F N = weight = mg = (150 kg)(9.8 N/kg) = 1470 N F Fr =  k F N = (.7)(1470) = 1029 N = 1000 N 1000 N W

 s =.62,  k =.48 What is the acceleration if there is a force of 72 N in the direction an 8.5 kg block is already sliding? F Fr =  k F N, F N = mg, F Fr =  k mg F Fr = (.48)(8.5 kg)(9.8 N/kg) = N F = ma = (8.5 kg)a, a = 3.77 = 3.8 ms m/s/s W 72 N 8.5 kg v F Fr

 s =.62,  k =.48 A 22 kg block is sliding on a level surface initially at 12 m/s. What time to stop? F Fr =  k F N, F N = mg, F Fr =  k mg F Fr = (.48)(22 kg)(9.8 N/kg) = N F = ma = (22 kg)a, a = ms -2 v = u + at, v = 0, u = 12, a = ms -2, t = 2.55 s = 2.6 s 2.6 s W 22 kg v=12m/s F Fr

 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? F Fr =  k F N, m = 6.5 kg F Fr = (.48)(6.5 kg)(9.8 N/kg) = N F = ma = (6.5 kg)(3.2 ms -2 ), F = = 51 N 51 N W F = ? F Fr 6.5 kg v a = 3.2 ms -2

 s =.62,  k =.48 A 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 = ms -2 F Fr =  k F N, F N = mg, F Fr =  k mg F Fr = (.48)(22 kg)(9.8 N/kg) = N F = ma = (22 kg)( ms -2 ), F = -22 N (left) -22 N (to the left) W 22 kg v=12m/s F Fr F = ?

 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? F Fr =  k F N, F N = mg, F Fr =  k mg F Fr = (.48)m(9.8 ms -2 ) = m(4.704 ms -2 ) F = ma = m(1.2 ms -2 ) 35 N = m(4.704 ms -2 ) + m(1.2 ms -2 ) = m(4.704 ms ms -2 ) m = (35 N)/(5.904 ms -2 ) = kg = 5.9 kg 5.9 kg W 35 NF Fr m v a = 1.2 ms -2

Page 3 - Inclined planes

 mg F perp = mgcos  F || = mgsin  F N = mgcos  (Causes friction) ( F Fr =  k F N ) 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 11 | 2 | 3 | 4 | TOC

Find F perp, F ||, the kinetic and maximum static friction: F || = mgsin(  ) = (3.52 kg)(9.8 N/kg)sin(42.0 o ) = N F perp = mgcos(  ) = (3.52 kg)(9.8 N/kg)cos(42.0 o ) = N F Fr(kinetic) =  k F N = (.37)(25.64 N) = 9.49 N F Fr(static) <  s F N = (.82)(25.64 N) = N 26 N, 23 N, 9.5 N, 21 N W  = 42.0 o 3.52 kg  s =.82  k =.37

No, it will not stay. The maximum static ( F Fr(static) <21.02 N ) friction is smaller than the gravity parallel to the plane ( F || =23.08 N) No, Blue W  = 42.0 o 3.52 kg  s =.82  k =.37 F || = N, F Fr(kinetic) = 9.49 N, F Fr(static) < 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  = 42.0 o 3.52 kg  s =.82  k =.37 F || = N, F Fr(kinetic) = 9.49 N, F Fr(static) < N What will be its acceleration down the plane if it is sliding down the plane? + - = (3.52 kg)a, a = m/s/s = -3.9 m/s/s N N

= (3.52 kg)a, a = m/s/s = -9.3 m/s/s v = 0, u = 5.0 m/s, a = m/s/s, v 2 =u 2 +2as, s = 1.35 m = 1.4 m -9.3 m/s/s, 1.4 m W  = 42.0 o 3.52 kg  s =.82  k =.37 F || = N, F Fr(kinetic) = 9.49 N, F Fr(static) < 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? N 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  = 42.0 o 3.52 kg  s =.82  k =.37 F || = N, F Fr(kinetic) = 9.49 N, F Fr(static) < N What force in what direction will make it accelerate and slide up the plane at 6.7 m/s/s? N N F = ? = (3.52 kg)(+6.7 m/s/s), F = N = 56 N

4.8 N W  = 42.0 o 3.52 kg  s =.82  k =.37 F || = N, F Fr(kinetic) = 9.49 N, F Fr(static) < 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 = (3.52 kg)(-2.5 m/s/s), F = N = +4.8 N N N F = ?

Page 4 - Statics

Force equilibrium: Step By Step: 1.Draw Picture 2.Calculate weights 3.Express/calculate components 4.Set up a = 0 equation for x and another for the y direction 5.Do math.

54 kg If F 1 is 185 N, what is F 2 ? F1F1 F2F 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: F 1 + F = 0, but since F 1 = 185 N 185 N + F = 0, so F 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 N F = ? 2.15 m5.82 m 1. Torque about the pivot point 2 and 3. (2.15 m)(5.25 N) - (5.82 m)F = 0 F = 1.94 N (mech. adv.)

Whiteboards Simple Torque Equilibrium 1 | 2 | 3 (One question like this)

315 N 87.5 N 12 mr = ? Find the missing distance. Torque about the pivot point. 43 m (315 N)(12 m) - (87.5 N)r = 0 r = 43.2 m = 43 m W

34 N F = ? 1.5 m6.7 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 = 0 F = 6.2 N W

512 N F = ? 2.0 m4.5 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) = 0 F = 362 N = 360 N W 481 N 3.1 m

Force Equilibrium: 1.Draw picture 2.Calculate weights 3.Draw arrows for forces. (weights of beams act at their center of gravity) 4.Make components 5.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 ACW r 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)

T 1 + T N N = 0 Step 1 - Set up your vertical force equation 52 kg 77 kg T1T1 T2T2 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. T 1 and T 2 are up, and the beam and person weights are down: Beam: -(52 kg)(9.8 N/kg) = N (down) Person: -(77 kg)(9.8 N/kg) = N (down) T 1 + T N N = 0

Nm Nm - (18.0 m)(T 2 ) = 0 Step 2 - Let’s torque about the left side Set up your torque equation: torque = rF 52 kg 77 kg T1T1 T2T2 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. T 1 = 0 Nm torque, (r = 0) Beam: (9.0 m)(509.6 N) = Nm (CW) Person: (13.0 m)(754.6 N) = Nm (CW) T 2 : T 2 at 18.0 m = -(18.0 m)(T 2 ) (ACW) Finally: Nm Nm - (18.0 m)(T 2 ) = m 13.0 m 18.0 m N N

T 2 = N T 1 =464.4 N Step 3 - Math time. Solve these equations for T 1 and T 2 : Nm Nm - (18.0 m)(T 2 ) = 0 T 1 + T N N = 0 52 kg 77 kg T1T1 T2T2 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 Nm Nm = (18.0 m)(T 2 ), T 2 = N T N N N = 0, T 1 = N