Chapter 4

Force and Motion

Younes Sina

Force may be defined as the cause of motion and deformation.When a force is applied to an object, the object either moves orchanges shape or both. In most cases, it is not possible to detect the deformation by naked eyes at the molecular or atomic level.Deformation occurs no matter how small.

The formula ΣF = M.a is called the "Newton's 2nd Law of Motion."

Example : An 850-kg car is accelerating at a rate of 2.4m/s2 to the right along a straight and horizontal road where it experiences an overall frictional force of 1500N. Determine the force that its engine exerts.

M= 850 kg

a= 2.4m/s2

Ff =1500N

Solution:

If "to the right" is taken to be positive as usual, Fe is positive, and Ff negative. Note that friction always opposes the direction of pending motion. Applying Newton's 2nd law:

ΣF = MaFe - Ff = MaFe - 1500N = (850kg)(2.4m/s2)Fe = 1500N + 2040N = 3500N

Note that the 2040N must be rounded to 2 significant figures and then added to the 1500N.

Example : A 2400-kg truck is moving at a constant speed of 15m/s on a horizontal and straight road that offers an overall frictional force of 1800N. Calculate (a) its acceleration(b) the engine force (c) the distance it travels in 35s(d) its acceleration if it changes its speed to 25m/s in 8.0 seconds,(e) the engine force in this case.

M= 2400 kg

v= 15 m/s

Ff =1800N

Solution:(a) Since the truck's speed is initially constant; thereforea1 = 0(b) ΣF = M.aFe - Ff = M.a1

Fe - 1800N = (2400kg)(0)Fe = 1800N + 0 = 1800N(c) x = v t (Equation of motion for constant velocity)x = (15m/s)(35s) = 530m (rounded to 2 sig. fig.)(d) a2 = (Vf - Vi) / ta2 = (25m/s - 15m/s) / 8.0s = 1.25 m/s2

(e) ΣF = M.aFe - Ff = M.a2

Fe - 1800N = (2400kg)(1.25m/s2)Fe = 1800N + 3000N = 4800N

Example : A car that weighs 14700N is traveling along a straight road at a speed of 108 km/h. The driver sees a deer on the road and has to bring the car to stop ina distance of 90 m. Determine (a) the necessary deceleration(b) the stopping force (c) the brakes force, if the road friction is 2100N,(d) the stopping time

W=14700 N v=108 km/h

Ff = 2100N

x=90 m

Solution:The mass of the car and its velocity in (m/s) must be determined first.w = M. g → M = w/gM = (14700N) / (9.8 m/s2) M = 1500kgVi = (108 km/h ) (1000m / 1km) ( 1h / 3600s) = 30.m/s(a) Vf

2 - Vi2 = 2 a x → (0) 2 - (30.)2 = 2 (a) (90.0m)

-900 = 180aa = - 5.0 m/s2 (deceleration)(b) ΣF = MaΣF = (1500kg)(- 5.0 m/s2 ) = -7500N(c) ΣF = Fbrakes + Ffriction

-7500N = Fbrakes - 2100NFbrakes = - 5400N(d) a = (Vf - Vi) / tt = (Vf - Vi) / at = (0 - 30.m/s) / (-5.0m/s2)t = 6.0 seconds

Example: A 20.0-kg crate is on a horizontal and frictionless surface as shown.(a) Calculate and show the vertical forces acting on this crate.(b) Knowing that the crate is being pushed to the left by a53-N force, what magnitude force (F) to the right must be appliedonto the block to give it an acceleration of 2.5m/s2 to the right?

W=20.0 kg F=53 N a=2.5 m/s2F

Example : An 80.0-kg man is standing in an elevator. Determine the force of the elevator onto the person if the elevator is (a) accelerating upward at 2.5m/s2

(b) going upward at constant speed (c) coming to stop going upward at a deceleration of 2.5m/s2

(d) going downward at an acceleration of 2.5m/s2.

Solution: The force of the elevator onto the person is nothing but the normal reaction, N, of the floor onto his feet. For each case, a force diagram must be drawn. Let's take the +y-axis to be upward.

(a) w = M.g = (80kg)(-9.8 m/s2) = -780NThis is the case that the elevator has just started going

upward. Since its speed has to change from zero to some value, it has to accelerate upward and the person feels heavier because the floor of the elevator exerts a normal reaction, N , onto the man that is greater than his weight. This creates a nonzero net force and therefore accelerates the person.ΣF = M.aN - Mg = M.aN =M (g + a)N =M (g + a) = (80.0kg)(9.8+2.5)m/s2 = 984N

(b) w = Mg = (80kg)(-9.8 m/s2) = -780NIn this case, since the elevator goes up at constant speed, its acceleration is zero and so is the acceleration of the man.Zero acceleration means zero net force acting on the man.This requires ( N ) to be equal to ( w ) in magnitude.ΣF = Ma ; N - Mg = Maa=0 →N =M.g = (80kg)(9.8)m/s2 = 780N

(c) w = Mg = (80kg)(-9.8 m/s2) = -780NIn this case, the elevator is coming to stop in its going upward. In other words, it decelerates as it goes upward. We all have this experience that during such slowing down, we feel lighter. We will notice that the magnitude of the normal reaction, N,

becomes less than that of w.ΣF = MaN - Mg = MaN =M (g + a)N =M (g + a) = (80kg)(9.8 - 2.5)m/s2 = 580N

(d) w = Mg = (80kg)(-9.8 m/s2) = -780NWhen the elevator starts going downward, its speed changes from zero to some value, and therefore it accelerates.This time we use -2.5 m/s2 because the acceleration vector is downward.ΣF = Ma ; N - Mg = Ma ; N =M (g + a)N =M (g + a) = (80kg)(9.8-2.5)m/s2 = 580N

Example : In the figure shown, determine the acceleration of the system of blocks:

2)If the cord is broken, does free fall occur?1) Does free fall occur?

"Yes""No"

Solution: Block B is not free to fall and must pull block A in addition to moving itself. The vertical forces on block A cancel each other, according to the Newton's 3rd law. The force of gravity on block B is the cause of motion while block A is on the horizontal surface. In fact, we are only interested to find the acceleration of the system of blocks while A slides horizontally. Let's summarize: (1) the force that causes motion is WB and (2) this force has to move both masses (MA + MB). Since the system is connected, both blocks move at the same acceleration (same magnitude).We may write:

ΣF = MawB = (MA + MB) a29N = ( 5.0 kg + 3.0 kg) aa = 3.6 m/s2

Example: Vehicles A and B are shown in nine different cases.In each case, a statement is made on the left. Refer to the figurein the middle and determine if the statement is true (T) or false (F).

1) MA = MB. If A is pushing B at constant velocity,then F1 = F2.

T

2) MA = MB; If A is pushing B and accelerating,then F1 >F2.

3) MA = MB; If A is pushing B and decelerating,then F1 <F2.

F

F

F1=MA. a

F2=MB. a

4) MA > MB; If A is pushing B at constant velocity,then F1 > F2.

5) MA > MB; If A is pushing B and accelerating,then F1 > F2.

F

F

6) MA > MB; If A is pushing B and decelerating,then F1 < F2.

7) MA < MB; If A is pushing B at constant velocity,then F1 = F2.

F

T

8) MA < MB; If A is pushing B and accelerating,then F1 > F2.

9) MA < MB; If A is pushing B and decelerating,then F1 < F2.

F

F

Friction: Friction is the result of engagement of surface

irregularities between two surfaces in contact.

Coefficient of Kinetic Friction ( μk ) on a horizontal surface: horizontally applied force (Fappl.)/weight of the object (w)

to slide the object at a constant velocity

Fappl. = Fk

w = N

Fk = μkN

Example : The coefficient of kinetic friction between a cement block and a plank of wood is 0.38 The block has a mass of 15 kg and is placed horizontally on the plank. Find the magnitude of the horizontal force that can push the block to the right at a constant velocity.

Example : In the figure shown, determine the magnitude of the horizontal force to the right that can move the block at (a) constant velocity(b) at an acceleration of 3.0 m/s2.

Solution: (a)

W= Mg = (25kg)(9.8m/s2) = 245 N

N = 245 N ; Fk = μkN = (0. 26)(245N) = 64 N

ΣF = Ma ; F - 64 = (25)( 0 ) ; F = 64 N

Solution: (b)

ΣF = Ma ; F - 64 = (25)(3. 0) N ; F = 139 N

Coefficient of Static Friction ( μs ) : On a horizontal surface, the ratio of thehorizontally applied force (Fappl.) to an object to the weight of the object (w),to bring the object onto the verge of slipping, is called the coefficient ofstatic friction. This is mathematically written as:

on the verge of slippingN = w

Fappl. = Fs

Example : The coefficient of static friction between a cement block and a plankof wood is 0.46. The block has a mass of 15 kg and is placed horizontally on theplank. Find the magnitude of the horizontal force that can bring the block ontothe verge of slipping.

Friction Laws:There are 5 laws for friction. The first 3 apply to the force of friction, and thelast 2 to the coefficient of friction.

1) Force of friction ( Ff ) is proportional to the coefficient of friction ( μ ) and the normal force ( N ).Fs = μsN

Fk = μkN

Note that:μs > μk → Fs > Fk

It takes a greater force to bring an object onto the verge of slipping than pushing it at a constant velocity.2) Force of friction ( Ff ) is always tangent or parallel to the contacting surfaces.3) Force of friction ( Ff ) always opposes the direction of pending motion.4) Coefficient of friction ( μ ) depends on the materials of the contacting surfaces.5) Coefficient of friction ( μ ) depends on the smoothness of the contacting surfaces.

Example : Two kids are sitting on the opposite sides of a 2.9 m long table and sliding a 150-gram empty cup toward each other, back and forth. The game isto give the cup the right initial velocity at one edge such that it comes to stop exactly at the opposite edge as shown. The diameter of the cup is 10.0cm.The coefficient of kinetic friction between the cup and the horizontal tabletopis 0.12. Determine the necessary initial speed.

X= 2.9 m Φ= 10.0cm

μ= 0.12

Solution:

We need to use the kinetic equation, ΣF = Ma in order to solve for acceleration.

What are the forces acting on the cup after it is given an initial instant push?

The only acting force is the force of kinetic friction, Fk.

To find Fk , we need to know N , and consequently (w).

w = Mg

w = Mg = (0.15kg)(9.8 m/s2) = 1.47N

N = 1.47N

Fk = μkN

Fk = (0.12)( 1.47 N ) = 0.176 N

ΣF = Ma

- 0.176N = (0.150kg)(a)

a = -1.173 m/s2

Vf2 - Vi

2 = 2 a x

02 - Vi2 = 2 (-1.173 m/s2)(2.9m)

Vi = 2.6 m/s

Example : A truck that weighs 29,400N traveling at 72.0 km/h on a horizontal and straight road skids to stop in 6.00s. Determine (a) its deceleration(b) the stopping force(c) the kinetic coefficient of friction between its tires and the horizontallystraight road,(d) the stopping distance.

V= 72.0 km/h W= 29,400N

t= 6.00 s

Solution:(a) a = (Vf - Vi) / ta = (0-20.0m/s) / (6.00s) = - 3.33 m/s2.w = Mg29,400N = M(9.8m/s2)M = 3,000kg

(b) ΣF = Ma0 - Fk = (3000kg)(- 3.33m/s2) = -10,000NFk = 10,000N

(c) Fk = μkN

10,000N = ( μk ) (29400N)μk = (10,000N) / (29400N) = 0.40

(d) x = (1/2)a t2 + Vi tx = (1/2)(-3.33m/s2)(6.00s)2 + (20.0m/s)(6.00s) = 60.0mVf

2 - Vi2 = 2 a x

(0)2 - (20.0m/s)2 = 2(-3.33m/s2) xx = 60.0m

Example : A 12-kg box is placed on a horizontal floor for which μs = 0.43and μk = 0.33. Does a 57-N force, applied horizontally to this box, put it intomotion? If yes, will the motion be accelerated or at constant speed?If accelerated, how far will it travel in 3.0s?

M= 12 kg

μs = 0.43μk = 0.33

F= 57N

Solution: We know that Fs > Fk .If Fs (the force of static friction) is less than 57-N, the horizontally applied force, motion will occur. Let's calculate Fs.w = Mgw = (12kg)(9.8 m/s2) = 118NN = 118NFs = μsN

Fs =(0.43)118N = 51NSince Fs < 57N ; therefore, motion occurs.Once motion occurs, μk takes over.Fk = μkN

Fk = (0.33)(118N) = 39NThe motion will be accelerated because the net force is not zero.ΣF = Ma57N - 39N = (12kg)( a )a = 1.5 m/s2

x = (1/2)a t2 + Vi tx = (1/2)(1.5m/s2)(3.0s)2 + (0)(3.0s)x = 6.8m

Example : In the figure shown, determine the magnitudeof force F that gives the block an acceleration of 1.75m/s2

on the horizontal surface to the right.

To solve for F and N, let's move these unknowns to the left of each equation while moving the known values to the right sides.

1) N - Fsin30 = 147N - 0.500 F = 1472) -.021N + Fcos30 = 26.25-0.21 N + 0.866 F = 26.25

Alternate Solution: Before rearranging, substitute for N from the 1st equation

into the 2nd one as shown:Fcos(30) - 0.21(147 + Fsin30) = 26.25now, there is only one unknown, F.0.866F - 30.87 - 0.105F = 26.250.761F = 57.12F = 75NFrom the N equationN = 147 + 75sin30N = 190 N

Homework:All problems in chapter 4