Chapter 6 : Do listed problems.  
Simulations
Quiz 6
5 6 7 14 15 17 18 19 21 29 37
40 41 42 46 51
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error log #1: Problem 42 uses the angle 10 degrees; the change has been made below.
error log #2: Problem 37 uses the symbol ay consistently; the change has been made below. 
In force problems, we must choose  positive x or y directions to compute the net force component along each axis. We can assign the positive direction  to be  direction of motion  to compute the acceleration component.  If, say,  you compute ax > 0, then the particle  speeds up along the x-axis.   If ax < 0,  the particle slows down.  If ax = 0, the  x-component of velocity is constant.    

Sometimes it's helpful to assign the direction of motion (i.e  the positive direction) to be the anticipated direction of the net force along the axis.  

5. If the  acceleration component > 0, then the particle  speeds up in that direction. If  acceleration component < 0,  the particle  slows down. 


(a)

X-Motion: positive direction to the right.
max = pos - neg
max = 4.0 (N) - 2.0 (N)
Y-Motion: positive direction up.
may = pos - neg
may = 3.0 (N) - 3.0 (N) = 0.
Given the mass m, find the x-directed acceleration ax .
(b)

X-Motion: positive direction to the right.
max = pos - neg
max = 4.0 (N) - 2.0 (N)
Y-Motion: positive direction up.
may = pos - neg
may = 3.0 (N) - 1.0 (N) - 1.0 (N) = 0.
Given the mass m, find the x-directed acceleration ax .

6.

(a)

X-Motion: positive direction along the force 5.0 (N) force component.
max = pos - neg
max = 5.0 - 3.0sin20 - 1.0
Y-Motion: positive direction along the  2.82 (N) force component.
may = pos - neg
may = 2.82 - 3.0cos20
Given the mass m, find the x and y-directed accelerations .  Please round off to two significant figures.

(b)
 
X-Motion: positive direction along the  3.0 (N) force component.
max = pos - neg
max = 3.0  - 2.0sin15 - 2.0cos15. (Look at the diagram carefully!)
Y-Motion: positive direction along the  1.414 (N) force component.
may = pos - neg
may = 1.414 - 2.0cos15 + 2.0sin15. (Look at the diagram carefully!)
Given the mass m, find the x and y-directed accelerations .  Please round off to two significant figures.
7. 

(a)

X-Motion: positive direction along the force component 5.0cos37 (N), rightward. 
max = pos - neg
max = 5.0cos37 - 2.0
Y-Motion: positive direction along the  2.0 (N) force component.
may = pos - neg
may = 2.0 + 5.0sin37 - 5.0
Given the mass m, find the x and y-directed accelerations .  Please round off to two significant figures.

(b)


X-Motion: positive direction along the  3.0 (N) force.
max = pos - neg
max = 3.0 + 5.0sin37 - 2.0
Y-Motion: positive direction along the 4.0 (N) force .
may = pos - neg
may = 4.0 - 5.0cos37
Given the mass m, find the x and y-directed accelerations .  Please round off to two significant figures.
14. 
(a) In the y - direction, let up be positive: N - mg = 0. Find N = mg.  In this case N = weight.
(b) On the moon, m remains the same. But the weight will be different since g is different: Again, N - mg = 0. Find N = mg.  In this case N = weight on the moon for a different value of g which you can Google or look up in the book. 
15. See class notes; the normal force equals the weight mg when the elevator is at rest   or when the upward velocity constant, parts a and c. In both these cases, the acceleration is zero. In part b, the normal force is greater than the weight mg when the elevator accelerates upward .  SEE  PAGES 161-2, FIG 6.10 AND "STOP TO THINK"  6.2. REPLACE FSP BY THE SYMBOL N, WHICH REPRESENTS  THE NORMAL FORCE.  BOTH FSP  AND N POINT UPWARD AND PLAY THE SAME ROLE MATHEMATICALLY. 
17. VELOCITY IS CONSTANT , SO ACCELERATION IS ZERO.  Units below are in Newtons (N)

max = pos - neg
0 = 385 + 350 - fk .
fk = µkN, where N = mg in this case.
Find the coefficient of kinetic friction.  
18.  Assume motion does not occur. Then
0 = pos - neg
0 = 800 - fs, where fs  is the static friction force.
In other words,  fs = 800, assuming motion does not occur.   
But  fs must be less than or equal to µsN. You should know by now what the normal force magnitude N is  for  horizontal placements. Check if fs  is less than or equal to µsN. If true, then motion does not occur. 

If false, then motion occurs, and you could find the acceleration using:

max = pos - neg
max = 800 - fk , where fk = µkN. 

Does motion occur?
19.
max = pos - neg
max = fs  - 0.  
Note that the static friction force causes the crate to accelerate without slipping on the accelerating belt !
More discussion of this subtle point later !
To get the maximum acceleration of the belt, set fs  = µsN. You should know by now what the normal force magnitude N is  for  horizontal placements.
21. 
max = pos - neg
max = 0 - neg
max = - µkN. The acceleration is negative. You should know by now what the normal force magnitude N is  for  horizontal placements. Find the acceleration theoretically, using symbols. Now, write: 

Vf2 = 0 = Vi2 +2ax·(65 m).  Substitute the acceleration,  then find the initial velocity.
29. Break up the problem in two dimensions, x and y. on both axes, the net force = 0. 

max = pos - neg
0 = T2sin30 - T1sin20. 
may = pos - neg
0 = T2cos30 + T1cos20 - mg.

Solve both equations for  tension magnitudes  T1 and T2.
37. You need to find the upward directed velocity when the object leaves the gun barrel. Here's how:  
First find the acceleration. 
may = pos - neg

may = 2.0 (N)  - mg

Then plug into:

Vf2 =  Vi2 +2ay·(1 m), where the initial velocity = 0. Find the final velocity  which corresponds to the end of the barrel.
 
The use:
Vff2 = 0 = Vf2  - 2g·h, where h is the distance above the end of the barrel. Note Vff  is  the zero velocity at the very top.  
40. 

(a)   Assume motion is just about to occur.  Then
0 = pos - neg
0 = T - fs, where fs  is the static friction force. But  fs must be  equal to µsN. You should know by now what the normal force magnitude N is  for  horizontal placements. Find T. The static friction coefficient is in Table 6.1. 
(b) After motion begins, use:
max = T - fk , where fk = µkN. Note: T = 20 (N). Find the acceleration , then use Chapter 2 methods to find the velocity after moving a distance 1.0 m.    Assume the initial velocity is zero. The kinetic friction coefficient is in Table 6.1 for steal-on-steel. 
(c) Repeat part (b) using the lower kinetic coefficient in Table 6.1 corresponding to steel-on-steel (lubricated).    
41. 

(a) max = F - fk , where fk = µkN and  F = thrust force magnitude. Find the acceleration, then use Vf =  Vi + ax·t, where  time t is given and the object starts from rest. 
(b) Find the distance D traveled during the  positive acceleration: Vf2 =  Vi2 +2ax·D, where object starts from rest. 
Then find the distance D' traveled during the negative acceleration while object slows down to zero: 
Vff2 =  0 =Vf2 +2ax'·D'  , where ax' = -µkN/m. Find the sum D + D'. 
42. The x -axis is parallel to the incline, the positive direction downward along the motion. Let the thrust force, pointing in the positive x-direction,  have magnitude F. Then: 
max = pos - neg
max = F + mgsin10 - fk.
Note:  fk = µkN. You should know by now the formula for the normal force  on an inclined plane. Hint: In the y-direction,  the acceleration is zero. Thus N = magnitude of component of gravitational force perpendicular to the surface.  Substitute this expression into: max = F + mgsin10 - fk.   Use the kinematical equation from Chapter 2: 
Vf2 =  Vi2 +2ax·L, where L is the distance moved along the incline. Sam starts from rest.  L is given through the trigonometric  relationship L = h/(sin10), where the inclined height h is given. Solve for ax . Then solve for the kinetic friction coefficient by substituting into: max = F + mgsin10 - µkN.  
46. 
(a) The x -axis is parallel to the incline, the positive direction upward along the motion.  Then: 
max = pos - neg
max = 0 - mgsin30 - fk.
Note:  fk = µkN. You should know by now the formula for the normal force  on an inclined plane. Hint: In the y-direction,  the acceleration is zero. Thus N = magnitude of component of gravitational force perpendicular to the surface.  Substitute this expression into: max = - mgsin15 - fk.  Solve for ax. The acceleration is negative. Then use the kinematical equation from Chapter 2: 
Vf2 =  Vi2 +2ax·L , where L is the distance moved along the incline. The initial velocity is given, the final velocity  zero. 
Find L. (See Table 6.1 for the kinetic friction coefficient.) The vertical height h is given through the trigonometric  relationship discussed in the previous problem. Find h. 
(b) The x -axis is parallel to the incline, but now the positive direction is downward along the motion.  Then: 
max = pos - neg
max = mgsin30 - fk.
Note:  fk = µkN. N = magnitude of component of gravitational force perpendicular to the surface.  Substitute this expression into: max = mgsin30 - µkN.  Solve for ax. The acceleration is positive. Then use the kinematical equation from Chapter 2: 
Vf2 =  Vi2 +2ax·L , where L is the distance moved along the incline. The initial velocity is zero. L is given in the previous part. Find the final velocity. Note: The final velocity will be smaller in magnitude  than  the  initial speed in the previous part. We will find later that the heat expended during the motion decreases the mechanical energy, thereby reducing the final speed from the initial value of  part (a). 
# 51 WILL BE DISCUSSED IN CLASS BUT HERE IS A LITTLE HINT:
ASSUME THE MOTION IS UPWARD; LET THE Y DIRECTION BE UP POSITIVE.

m*ay = pos - neg = F*sin45 - mg - uk*N, where F = Fpush shown.

N is the normal force; uk the coefficient of kinetic friction.

IN THE X DIRECTION, LET RIGHTWARD BE POSITIVE :

m*ax = 0 = pos - neg = N - F*cos45.

Now, m*ay = 0 because the velocity is constant, whether the box moves 
down or up. We move *up* in this example. Solve the two equations for F and N.

NOW,  THE BOOK ASSUMES THE MOTION IS DOWN; HERE IS THE HINT FOR THAT PROBLEM:

In that case, the positive y-direction is down, and the positive
x-axis is rightward.

m*ay = pos - neg =  mg - F*sin45 -  uk*N, where F = Fpush shown.

N is the  normal force; uk the coefficient of kinetic friction.

m*ax = 0 = pos - neg = N - F*cos45.

Now, m*ay = 0 because the velocity is constant. Solve the two
equations for F and N.
error log #1: Problem 42 uses the angle 10 degrees; the change has been made below.
error log #2: Problem 37 uses the symbol ay consistently; the change has been made below.