Chapter 4

Force and Motion

Force may be defined as the cause of motion and deformation.  When a force is applied to an object, the object either moves or changes 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.

In Chapter 1, force was defined as the product of mass and acceleration.  Simply F = Ma.  A more useful form of this formula is ΣF = Ma.   ΣF means the sum of forces acting on mass M.  Since forces acting on an object may act in opposite directions, ΣF is also called the net force.

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

For example, a car moving along a straight and horizontal highway, experiences an engine force Fe, while being opposed by an overall frictional force, Ff ( road friction as well as air resistance).  If the car is moving to the right and to the right is taken to be the positive direction, then Fe acts to the right and Ff acts to the left.  The net force is ΣF = Fe - Ff .

Example 1: 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.

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 = Ma    ;    Fe - Ff  =  Ma    ;    Fe - 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 2: 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, and (e) the engine force in this case.

Solution:  (a) Since the truck's speed is initially constant; therefore,  a1 = 0.

(b)    ΣF = Ma    ;    Fe - Ff  =  Ma1       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) / t    ;    a2  = (25m/s - 15m/s) / 8.0s  =  1.25 m/s2

(e)  ΣF = Ma   ;   Fe - Ff  =  Ma2   ;   Fe - 1800N  =  (2400kg)(1.25m/s2)   ;   Fe  = 1800N  + 3000N  =  4800N

Example 3:  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 in a distance of 90.m.  Determine (a) the necessary deceleration, (b) the stopping force, (c) the brakes force, if the road friction is 2100N, and (d) the stopping time.  Try to solve the problem yourself before looking at the solution.

Solution:  The mass of the car and its velocity in (m/s) must be determined first.  Recall that w = Mg.   The mass of the car is therefore,  M = w/g,    or    M = (14700N) / (9.8 m/s2) ,    or    M = 1500kg.

Vi = (108 km/h ) (1000m / 1km) ( 1h / 3600s) = 30.m/s    (Use horizontal fraction bars only)

(a)   Vf2 - Vi2 = 2 a x    ;    (0) 2 - (30.)2 = 2 (a) (90.0m)    ;    -900 = 180a    ;    a = - 5.0 m/s2 (deceleration).

(b)    ΣF = Ma    ;     ΣF = (1500kg)(- 5.0 m/s2 ) = -7500N

(c)    ΣF = Fbrakes + Ffriction    ;    -7500N = Fbrakes - 2100N    ;    Fbrakes = - 5400N.

(d)    a = (Vf - Vi) / t    ;    t = (Vf - Vi) / a    ;    t = (0 - 30.m/s) / (-5.0m/s2)    ;    t = 6.0 seconds

Newton's Laws:

1) If an object is under a zero net force, it is either stationary or if moving, it moves at constant velocity.  Note that constant velocity means constant speed plus constant direction that means along a straight line.

2) A nonzero net force ΣF acting on mass M causes an acceleration a in it such that ΣF = Ma.  The acceleration has the same direction as the applied net force.

3) There is a reaction for every action, equal in magnitude, but opposite in direction.

An example covering the 1st and 2nd laws has already been made (Example 2).  For the 3rd law, look at the following example:

Example 4: 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 a 53-N force, what magnitude force (F) to the right must be applied onto the block to give it an acceleration of 2.5m/s2 to the right?

Note that w = 196 N must be rounded to 2 sig. fig. and written as w = 2.0x102 N.  The same is true for F = 103N that must be rounded to 100 and written as F = 1.0x102 N.

Example 5:  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, and (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.  Make sure that you carefully draw each force diagram with minimal looking at the following force diagrams.

 (a)  w = Mg = (80kg)(-9.8 m/s2) = -780N    This 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 = Ma ;   N  - Mg = Ma ;   N  =M (g + a) N  =M (g + a) = (80.0kg)(9.8+2.5)m/s2 = 984N (b)   w = Mg = (80kg)(-9.8 m/s2) = -780N     In 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 = Ma ;   N  =M (g + a) N  =M (g + a) = (80kg)(9.8+0)m/s2 = 780N (c)  w = Mg = (80kg)(-9.8 m/s2) = -780N     In 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 = Ma ;   N  - Mg = Ma ;   N  =M (g + a) N  =M (g + a) = (80kg)(9.8 - 2.5)m/s2 = 580N (d)   w = Mg = (80kg)(-9.8 m/s2) = -780N     When 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 6:   In the figure shown, determine the acceleration of the system of blocks:

Solution:  Does free fall occur?  If the cord is broken, does free fall occur?  The answer to the 2nd question is definitely "Yes."  The answer to the first question is now more clear, "No."  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 = Ma    ;    wB = (MA + MB) a    ;    29N = ( 5.0 kg   +   3.0 kg) a    ;    a = 3.6  m/s2

Example 7:

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

 MA = MB.  If A is pushing B at constant velocity,                            then  F1 = F2. ( T )  or ( F ) ? MA = MB;  If A is pushing B and accelerating,                            then F1 >F2. ( T )  or ( F ) ? MA = MB;  If A is pushing B and decelerating,                             then F1 MB;  If A is pushing B at constant velocity,                             then F1 > F2. ( T )  or ( F ) ? MA > MB;  If A is pushing B and accelerating,                              then F1 > F2. ( T )  or ( F ) ? MA > MB;  If A is pushing B and decelerating,                              then F1 < F2. ( T )  or ( F ) ? MA < MB;  If A is pushing B at constant velocity,                               then F1 = F2. ( T )  or ( F ) ? MA < MB;  If A is pushing B and accelerating,                               then F1 > F2. ( T )  or ( F ) ? MA < MB;  If A is pushing B and decelerating,                               then F1 < F2. ( T )  or ( F ) ?

Chapter 4 Test Yourself 1:

1) Force may be defined as (a) the cause of motion  (b) the cause of deformation  (c) both a and b.

2) In ΣF = MaF and a  (a) are both scalars  (b) are both vectors  (c) form a scalar-vector pair.    click here

3) The direction of F and a in  ΣF = M (a) are the same (b) are different  (b) are normal to each other.

4) In mathematics and physics, the word "normal" means (a) parallel  (b) perpendicular  (c) not abnormal.    click here

5)  ΣF = Ma implies (a) the proportionality of the net force and the acceleration it generates in mass M  (b) the proportionality of the net force and mass  (c) both a and b.    click here

6) If the same force is applied to masses M1 and M2 separately, knowing that M1>M2,  then (a) M1 accelerates greater than M2 does  (b) M1 accelerates less than M2 does   (c) both M1 and M2 gain the same acceleration.    click here

7) The acceleration of gravity on an object in the vicinity of the Earth (a) is the same whether the object is stationary or falling freely  (b) acts on the object only if it is falling  (c) acts on the object when it is placed on the horizontal surface only.

8) The acceleration of gravity (a) is exactly zero where the Space Station is  (b) is not exactly zero where the Space Station is, but its direction is different there (c)  does not completely diminish with distance.    click here

9)  The direction of the acceleration (and hence, force of gravity) on an object around the Earth (a) always passes through the center of the Earth  (b) is perpendicular to the local free water surface  (c) both a and b.

10) The direction of g , the gravity acceleration, at any location is (a) the same as the plumb-line at that location  (b) is a few degrees different from the plumb-line at that location  (c) is normal to the plum-line at that location.    click here

11) According to Newton's 1st law (a) net force and mass are proportional  (b) a moving object under a zero net force has a constant speed only  (c) a moving object under a zero net force has a constant velocity.

12) When you are pushing a shopping cart horizontally and at a constant velocity, your force on the cart is (a) greater than the force you feel from the cart on your hand  (a) is less than the force you fell from the cart on your hand  (c) equal to the force you feel from the cart on your hand.    click here

13) When you are accelerating a shopping cart horizontally, your force on the cart is (a) greater than the force you feel from the cart on you  (a) is less than the force you fell from the cart on you  (c) equal to the force you feel from the cart on you.

14) When you are slowing down a shopping cart horizontally, your force on the cart is (a) greater than the force you feel from the cart on you  (a) is less than the force you fell from the cart on you  (c) equal to the force you feel from the cart on you.

15) Questions 12, 13, and 14 may be answered using (a) Newton's 1st law  (b) Newton's 2nd law  (c) Newton's 3rd law.

16)  Is the direction of  ΣF in ΣF = Ma, the same as the direction of acceleration (a)?  Ans:  ......  click here

17) If the net force acting on a mass (M) is not zero, can that mass move at constant velocity?  Ans.:  .......

18) If the net force on an object is zero, is the object definitely moving at constant velocity?   Ans.:  .......

19) A box weighs 35N and is placed on a horizontal surface.  Is the normal force 35N?   Ans.:  ......   click here

20)  A 75-N box is placed on a horiz. surf. with a boy sitting on it.  Is the normal force on the box 75N?   Ans.:  ......

21) When an object is accelerating horizontally on a horizontal surface, is the horizontal external force on the object equal to the horizontal friction force of surface on the object? Ans.:  ......   click here

22)  Does a lady standing in an elevator accelerating upward feel heavier because the normal force from the floor on her is greater than her weight?   Ans.:  ......

23) A spaceship in outer space is moving at constant velocity of 55 miles/sec.   An astronaut looking through a window notices a bolt becoming detached from the ship.  How many miles behind the ship will the bolt be in an hour?  Ans.:  ...... click here

24) A spaceship moving along a straight path wants to change its speed from 55 mi/s to 75mi/s at a relatively low acceleration of (1/3)g to keep the occupants in some relative comfort.  How long will it take?  How many miles does it travel for the change to happen.  Ans.:  ......

25) If the spaceship in (Question 24) weighs 490,000N on Earth, what average force should its jets exert on it during the acceleration period?  Ans.:  ......   click here

Friction:

Friction is the result of engagement of surface irregularities between two surfaces in contact.

Coefficient of Kinetic Friction ( μk ) : On a horizontal surface, the ratio of the horizontally applied force (Fappl.) to an object to the weight of the object (w), to slide the object at a constant velocity, is called the coefficient of kinetic friction.  This is mathematically written as:

Note that  Fappl. =  Fk and  w  =  N .

It is easy to understand that Fk ( force of kinetic friction) is the force that the horizontal surface exerts on the object and Fappl. is the force that equals  Fk.   If  Fappl.  is exactly equal to  Fk, the object slides at constant velocity.  We also know that if w and N are the only forces in the vertical direction, then N  = w ; therefore, the above formula may be written as

This formula is often written in its cross-multiplied form:  Fk = μkN .   This means that, if the coefficient of friction between two surfaces is known, and we can also determine the normal or compressive force between those two surfaces, we are then able to determine the necessary force that can slide one surface against the other at a constant velocity.

Example 8: The coefficient of kinetic friction between a cement block and a plank of wood is 0.38.  The block has a mass of 15kg 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 9:

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

 Solution: (a)   w = Mg = (25kg)(9.8m/s2) = 245N   N = 245N  ;  Fk = μkN  = (0. 26)(245N) =  64N   ΣF = Ma   ;   F - 64 = (25)( 0 )   ;   F = 64N    Solution: (b)    ΣF = Ma   ;   F - 64N = (25)(3. 0)N   ;   F = 139N

Coefficient of Static Friction ( μs ) : On a horizontal surface, the ratio of the horizontally 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 of static friction.  This is mathematically written as:

Again, N  = w , and Fappl. =  Fs , on the verge of slipping, and the above equation may be written as:

This formula is often written in its cross-multiplied form:  Fs = μsN .   This means that, if the coefficient of static friction between two surfaces is known, and we can also determine the normal or compressive force between those two surfaces, we are then able to determine the necessary force that can bring the object onto the verge of slipping.

Example 10: The coefficient of static friction between a cement block and a plank of wood is 0.46.  The block has a mass of 15kg and is placed horizontally on the plank.  Find the magnitude of the horizontal force that can bring the block onto the verge of slipping.

Friction Laws:

There are 5 laws for friction.  The first 3 apply to the force of friction, and the last 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        and      Fk = μkN.            Note that:    μs >  μk    ; thus,    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 11: Two kids are sitting on the opposite sides of a 3.0-m long table and sliding a 150-gram empty cup toward each other, back and forth.  The game is to 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 tabletop is 0.12.  Determine the necessary initial speed.

 Solution:  Time is not given.  At a first glance, you may think that  ( Vf2 - Vi2 = 2ax ) is a good idea. In this equation, the acceleration (a) of the cup is not known.  This means that we need to use the kinetic equation, ΣF = Ma in order to solve for acceleration, first.  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).   Therefore, we must start from w = Mg.    w = Mg = (0.15kg)(9.8 m/s2) = 1.47N ;     therefore,      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 - Vi2 = 2 a x     ;      02 - Vi2 = 2 (-1.173 m/s2)(2.9m)     ;    Vi = 2.6 m/s.

Example 12: 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 horizontally straight road, and (d) the stopping distance.  Important: First draw a diagram for the problem and show all forces acting on the truck.    g = 9.8m/s2.

Solution:  (a)    a = (Vf - Vi) / t    ;    a = (0-20.0m/s) / (6.00s) = - 3.33 m/s2.

w = Mg    ;    29,400N = M(9.8m/s2)    ;    M = 3,000kg.

(b)     ΣF = Ma    ;    0 - Fk =  (3000kg)(- 3.33m/s2)  = -10,000N    ;    Fk = 10,000N

(c)    Fk = μkN    ;    10,000N = ( μk ) (29400N)    ;    μk = (10,000N) / (29400N) = 0.40

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

or,   Vf2 - Vi2 = 2 a x    ;    (0)2 - (20.0m/s)2 = 2(-3.33m/s2) x    ;    x = 60.0m

Example 13:  A 12-kg box is placed on a horizontal floor for which  μs = 0.43 and μk = 0.33.  Does a 57-N force, applied horizontally to this box, put it into motion?  If yes, will the motion be accelerated or at constant speed?  If accelerated, how far will it travel in 3.0s?

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 = Mg    ;    w = (12kg)(9.8 m/s2) = 118N    ;    N  = 118N    ;        Fs = μsN     ;    Fs =(0.43)118N = 51N

Since Fs < 57N ; therefore, motion occurs.  Once motion occurs, μk takes over.

Fk = μkN    ;    Fk = (0.33)(118N) = 39N    ;    The motion will be accelerated because the net force is not zero.

ΣF = Ma   ;    57N - 39N = (12kg)( a )    ;    a = 1.5 m/s2    ;    the distance traveled will be

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

Example 14: In the figure shown, determine the magnitude of 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 = 147                  ;               N   -  0.500 F  =  147      ;

2)       -.021N  + Fcos30 = 26.25   ;     -0.21 N   +  0.866 F  =  26.25   ; Use a calculator

Alternate Solution:  Before rearranging, substitute for N  from the 1st equation into the 2nd one as shown:

Fcos(30) - 0.21(147 + Fsin30) = 26.25    ;    now, there is only one unknown, F.

0.866F - 30.87 - 0.105F = 26.25    ;      0.761F =  57.12    ;    F =  75N

From the N  equation,    N  = 147 + 75sin30    ;    N  = 190 N

Chapter 4 Test Yourself 2:    ( In answering the following, a force diagram is always very helpful.)

1) Friction is the result of the (a) engagement of surface irregularities between two contacting objects  (b) the molecular attraction between two contacting objects  (c) both a and b.     click here

2) The importance of normal force, N , between two contacting surfaces is that (a) it helps us calculate the force of friction  (b) it is a measure of the extent two objects (in contact) push against each other  (c) both a and b.     click here

3) It takes 12N to horizontally push a 45-N block, placed on a horizontal surface, at constant velocity.  Another 45-N block is placed on the top of the 1st one.  To push both blocks at constant velocity,  it takes a force of (a) 12N  (b) 24N  (c) 36N.

4) Coefficient of friction, μ , is (a) the ratio of force of friction, Ff,  to the normal force, N   (b) the ratio of the horizontally applied force, Fappl., to the weight force, w, when the object is on a horizontal surface moving at a constant velocity  (c) both a and b.

5) The formula for force of friction is (a) Ff = μ N     (b) Ff = μ /N    (b) neither a, nor b.     click here

6) On a horizontal surface, for a single block of weight w, and a horizontally applied external force, Fappl., on the block, if the block is sliding at constant velocity, we may write:    (a) N =  w     (b)Fappl.= Ff     (c) both a and b.    First draw a force diagram, then solve and answer.     click here

7) On a horizontal surface, it takes a horizontal force of (a)14N  (b)22N  (c)11N to push a 55-N block at a constant velocity, knowing that the coefficient of kinetic friction between the block and the surface is 0.40.  First draw a force diagram, then solve and answer.

8) On a horizontal surface, it takes a horizontal force of (a)98N  (b)22N  (c)11N to push a 50-kg block at a constant velocity, knowing that the coefficient of kinetic friction between the block and the surface is 0.20.  First draw a force diagram, then solve and answer.     click here

9) First draw a force diagram, then solve and answer.  To accelerate a 25-kg block horizontally on a horizontal surface at a rate of 4.0m/s2 , knowing that μk = 0.30, a horizontally applied force of (a)173.5N  (b)73.5N  (c) .5N is needed.

10) First draw a force diagram, then solve and answer.  To accelerate a 50-kg block horizontally on a horizontal surface at a rate of 4.0m/s2 , knowing that μk = 0.30, a horizontally applied force of (a)347N  (b)473.5N  (c)273.5N is needed.

 11)  In the figure shown, the applied force, F, is not horizontal, we may write:(a) N =  w   (b) N >  w    (c) N <  w    click here 12)  In the figure shown, the applied force, F, is not horizontal, we may write:(a) N =  w   (b) N >  w   (c) N <  w     click here

13) In the figure of Question 11, if F = (120N, -30.0º), then (a) N = 147N    (b) N = 60N    (c) N = 207N.

14) In the figure of Question 12, if F = (120N, +30.0º), then (a) N = 60N    (b) N = 87N    (c) N = 407N.

15) Redraw the figure of Question 11 with F = 140N , M = 23kg, and μk = 0.30 and calculate the acceleration of the block in its sliding to the right.  The acceleration is (a) 2.42m/s2    (b)1.42m/s2      (c) 5.42m/s2.    click here

16) Redraw the figure of Question 11 with F = 110N , M = 33kg, and μk = 0.150 and calculate the acceleration of the block in its sliding to the right.  The acceleration is (a)1.167m/s2     (b)1.47m/s2      (c)1.38m/s2.    click here

17) The coefficients of kinetic and static friction for a block and surface are 0.26 and 0.38 respectively.  If the block weighs 120N and the surface is horizontal, can a 35N horizontal force put the block into motion?  .......... click here

18)   The coefficients of kinetic and static friction for a block and surface are 0.36 and 0.54, respectively.  If the block weighs 325N and the surface is horizontal, can a 205N horizontal force put the block into motion?  .......... click here

19) If the answer for Question 18 is "Yes", at what acceleration will the block slide?  .......... click here

20) The coefficient of friction, μ, between two surfaces with constant characteristics (a) is proportional to the applied force  (b) is a constant  (c) depends on the materials of the contacting surfaces  (d) b and c.    click here

21) The force of friction, Ff,  (a) is proportional to the applied force (b) is a constant  (c) does not depend on the material of the contacting surfaces  (d) is proportional to the normal force, N .   click here

22) The coefficient of friction, μ,  (a) depends on the roughness or smoothness of the contacting surfaces  (b) is a constant  (c) depends on the material of the contacting surfaces  (d) all of the above.  click here

23) Suppose that you have a bucket of water and a string just strong enough to hold it hanging.  What happens if you try to lift the bucket by a few inches?  Does it (a) pull? or (b) break?

24) Refer to Example 6 and replace the 3.0-kg hanging block by a 5.0-kg one.   In the absence of friction, the system of blocks move at an acceleration of (a) 9.8m/s2   (b) 4.9m/s2    (c) 2.5m/s2.     click here

25) Draw a force diagram for the hanging block of Question 24 and show the 49-N force, the block's weight, as well as the unknown tension, T, of the cord.   T should be shown as an upward force vector on the top edge of the square you draw as the block.  Since you know the acceleration of the block from Question 24, you can apply ΣF = Ma  just to this block and solve for T, the tension in the cord.  We know that the block accelerates downward at 4.9 m/s2 as you have already solved.  If you assume upward to be positive, then T is positive, the 49N is negative and so is the acceleration.  Apply ΣFy = May and solve for T.  The answer is (a)17N   (b) 24.5N   (c)73.5N.

Problems:

1) A car traveling at a constant velocity on a horizontal and straight road is facing air resistance as well as a road frictional force of 2400N.  The engine is exerting a force of  4200N.  Calculate the force due to air resistance.

 2) In the figure shown, the 3.0-kg  hanging block causes motion of the two connected blocks at constant velocities.  Find the coefficient of friction between the top block and the horizontal table. 3) In the figure shown, find the coefficient of friction if the acceleration of the motion of the system of blocks has a magnitude of 0.75m/s2.

4) A 2.0-kg rock is hanging from a rope that can withstand a maximum tension of 33N.  (a) If the rope is pulled upward at an acceleration of 5.0m/s2, will the rope break?  (b) Will it break if the upward acceleration given to the rope is 7.0m/s2?  Support your answers with complete calculations.

5) A 55-kg lady is standing in an elevator.  Assuming g = 9.80 m/s2, calculate the force of elevator on her feet for the following cases:   the elevator is (a) at rest,  (b) moving downward at an acceleration of 3.0m/s2,  (c) moving down at a constant velocity,  and (d) is coming to stop in its downward motion at a deceleration of 3.0m/s2.

6) An 800.-kg car traveling at a velocity of 25 m/s eastward, comes to stop within a distance of 50.0m.  If the brakes apply a force of 3800N, find the road's overall frictional force.

7) A train moving at 30.0 m/s resumes this speed for 25.0 minutes.  Find (a) its acceleration and (b) the distance it travels during this period.   It then slows down to 20.0m/s within a distance of 125m.  Find (c) its deceleration, and (d) the elapsed time during the slowing down phase.

Answers: 1)1800N    2) 0.60    3) 0.54    4) No, Yes    5) 539N, 374N, 539N, 704N    6) 1200N    7) 0 , 28.0 miles, -2.00m/s2, 5.00s.