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.  SimplyF = 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, Fe is positive.    Ff that opposes motion and acts to the left is therefore negative.  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 its engine force.

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

Fe- Ff = Ma   ;    Fe-1500N = 850kg(2.4m/s2)    ;    Fe= 1500N+2000N = 3500N.

Example 2: A 2400-kg truck is moving to the right 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 velocity is initially constant; therefore a1 = 0.

(b) ΣF = Ma ;   Fe - Ff  =  Ma1;  Fe-1800N = 2400kg(0) ;   Fe =1800N + 0 =1800N.

(c) With a1= 0,   x = (1/2)a1t2 + vit  becomes  x = vt   or   x =15m/s(35s) = 530m.

(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 90m.  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 M of the car and its velocity in m/s must be calculated first. Since w = Mg; therefore,  M = w/g ;   M = 14700N/(9.8 m/s2) ;   M = 1500kg.

Vi = (108 km/h ) = (108000m /3600s) = 30m/s.   Use horizontal fraction bars only.

(a) Vf2 - Vi2 = 2ax   ;   02-302 = 2a(90m)  ;  -900 = 180a  ;  a = -5.0 m/s2.

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

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

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

Newton's Laws:

1) If an object is under a zero net force, it is either at rest 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 accelerates mass M at rate a 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.

The above examples were on Newton's 1st and 2nd laws.  For Newton's 3rd law, look at the following example:

Example 4:  A 20.0-kg crate is on a horizontal and frictionless surface as shown.  (a) Find and show the vertical forces acting on it.  (b) If a 53-N force is pushing it to the left, what force F must be applied to it to accelerate it at 2.50m/s2 to the right?  g = 9.80m/s2.

Solution:

Example 5:  An 80kg 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 in 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 elevator 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) = (80kg)(9.8+2.5)m/s2 = 980N. (b)  In this case, since the elevator goes up at constant velocity, 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)  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) 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 each block if (a) the string is cut, and (b) if only the stopper is removed.

Solution:

(a) If the string is cut:   B falls freely because with the string cut, B doesn't have to pull or move A. The acceleration magnitudes are:   aA = 0    and      aB = 9.8m/s2.

(b) If only the stopper is removed:

B can't fall freely and must pull A as well; therefore, its acceleration will not be 9.8m/s2.   According to Newton's 3rd law, the vertical forces on A cancel each other.  See the right side figure.  The force of gravity on B is the only cause of motion of the two blocks.  In fact, we are only interested to find the magnitude of acceleration of the system of blocks.  Let's summarize:  (1) the force that causes motion is wB and (2) this force has to move both: MA+MB.  Since the system is connected, both blocks move at the same acceleration magnitude.  Although both blocks move at the same acceleration magnitudes; however,  A accelerates horizontally while B accelerates vertically.   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 = Ma F and a  (a) are both scalars  (b) are both vectors  (c) form a scalar-vector pair.    click here.

3) The directions of F and a in ΣF = Mare (a) the same  (b) different  (c) 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 force of gravity on an object around the Earth (a) always passes through the center of the Earth  (b) is perpendicular to the local 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  (b) is less than  (c) equal to the force of 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 (b) is less than  (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  (a) less than  (c) equal to the force you feel from the cart on you.

15) Questions 12, 13, and 14 answers are based on Newton's (a) 1st law   (b) 2nd law   (c) 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 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 a constant velocity?   Ans.:  .......

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

20)  A 75-N box is placed on a horizontal surface 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.

The 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), if sliding at a constant velocity, is called the "coefficient of kinetic friction."   This is mathematically written as:

Note that whoever tries to pull the block to the right will face a resistant frictional force Fk as shown below.   For constant velocity motion, we may write:

Fk = Fappl.  and as we already know, magnitude wise,  N  = w.

It is easy to see that Fk ( force of kinetic friction) is the force that the horizontal surface exerts on the object and Fappl. is the force that must equal  Fk for constant velocity motion of the block.   If  Fappl.  is exactly equal to Fk, the object slides at a 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 also know the normal force, we can easily calculate the force of friction.  From this force of friction, we can then determine the force that is needed to slide one surface against another at a constant velocity.

Example 8: A 15kg block is placed on a horizontal plank plank of wood.  The coefficient of kinetic friction between the block and the wood plank is 0.38.   Find the magnitude of the horizontal force that can push the block on the plank to the right at a constant velocity.

 Solution: w = Mg = (15kg)(9.8m/s2 ) = 147N. This means that N = 147N as well. Fk = µk N = (0.38)(147N) = 56N. For constant speed motion, Fappl. must equal Fk = 56N, as shown in the diagram. Fappl. = 56N as well.

Example 9:

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

 Solution: Refer to the right figure.  (a)  w = Mg = (25kg)(9.8m/s2) =  245N.  N = 245N ;  Fk = μkN  = (0. 26)( 245N) =  64N.   ΣF = Ma ;  F-64 = 25(0) ;  F = 64N.  (b)  ΣF = Ma;     F - 64N = 25(3.0)N   ;   F = 139N.

The Coefficient of Static Friction (μs):   On a horizontal surface, the ratio of the horizontally applied force (Fappl.) that can bring an object onto the verge of sliding to the  weight of the object (w), to  is called " the coefficient of static friction."  This is mathematically written as:

Again, as shown below, N  = w , and on the verge of slipping, we must have Fappl. =  Fs .     The above equation may then 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 15kg block and a horizontal plank of wood that supports it is 0.46.   Find the magnitude of the horizontal force that can bring the block onto the verge of sliding.

Solution:

 w = Mg = (15kg)(9.8m/s2) = 147N. This means that N = 147N as well. Fs = µsN = (0.46)(147N) = 68N.On the verge of sliding, Fappli. must equal Fs = 68N, as shown in the diagram. Fappli. = 68N as well.

Friction Rules:

There are 5 rules 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 always tangent or parallel to the contacting surfaces.

2) Force of friction Ff always opposes the direction of pending motion.

3) Force of friction Ff is proportional to the normal force N .  The proportionality constant is nothing but the coefficient of friction.

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 sliding than pushing it at a constant velocity when in 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 3.00-m long table and sliding a 0.150 kg empty cup toward each other back and forth.  The game is to give the cup enough initial velocity at one edge of the table such that it comes to stop exactly at the opposite edge as shown.  The top view shown below.  The diameter of the cup is 10cm.  The coefficient of kinetic friction between the cup and the horizontal tabletop is 0.120.  Determine the necessary initial speed.

 Solution:  Time is not given.   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.  What horizontal forces do act on the cup after it is given an initial push?  The only horizontal force that acts on the cup 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 = (0.150kg)(9.81 m/s2) = 1.47N ;   therefore,   N  = 1.47N.   Fk = μkN     ;    Fk = 0.120 (1.47N) = 0.177 N.   ΣF = Ma    ;    - 0.177N = (0.150kg)(a)      ;         a = -1.18 m/s2.  Vf2 - Vi2 = 2ax  ;   02 - Vi2 = 2 (-1.18 m/s2)(2.90m)  ;  Vi = 2.62 m/s.

Example 12: A truck that weighs 29,400N is 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 coefficient of kinetic 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 = -3.33m/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 = 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.

Example 13:  A 12-kg box is placed on a horizontal floor for which  μs = 0.43 and μk = 0.33.  Does a 57N 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,  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 kicks in.

Fk = μkN    ;    Fk = 0.33(118N) = 39N.     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 is

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.

Solution:

 F is not horizontal, and therefore it has both horizontal and vertical effects.  The force diagram for the block is shown below. w = Mg = 15kg(9.8m/s2) = 147N. In the y-direction:  N = 147 + Fsin(30°).     (1) In the x-direction,   Fcos(30°) - 0.21N  = 15kg(1.75m/s2).            (2) The above two equations (solved simultaneously) give the values for F and N .

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)       -0.21N  + 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.

A Good Link To Try:

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) 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 a constant velocity.  Another 45-N block is placed on the top of the 1st one.  To push both blocks at a 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., to 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 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) 0.5N 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 motion 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 motion 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 a, b, and c.  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?   It (a) pulls   (b) breaks.

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 or 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 already 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.  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 friction 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 800kg car traveling at a velocity of 25 m/s eastward comes to stop within a distance of 50m.  If the brakes apply a force of 3800N, find the road's overall friction 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.