Site hosted by Angelfire.com: Build your free website today!
Main Desktop Lessons Desktop Review Desktop Lab Desktop Chart Desktop

    
Newton's Law Problems

Table of Contents
Problems
Answers

   
Problems

1. Lauren(mass = 55.0 kg) is laying on a sofa.   Table of Contents     Answer to 1.
(a)   What is the FN applied by the sofa?
(b) What is FN if Allie pushes down on her with a force of 300. N?
(c) What is FN if Allie is pulling up on her with a force of 300. N?
(d) What is FN if Allie is pulling up on her with a force of 600. N?
(e) What is the acceleration for each of the above?

  

2.  Stan(mass = 65.0 kg) is being pulled by Melissa with a force of 600. N and Jamel with a force of 900. N. There is an angle of 35.0o between the two. What happens to Stan?   Table of Contents    Answer to 2.

  

3. Tyler(mass = 75.0 kg) is swimming in the ocean. His kick and stroke combine for a forward force of 60.0 N. A current creates a force of 30.0 N and is coming from behind him at an angle 25o to the left. What is his velocity 15.0 seconds after he starts to swim. Assume the water resistance does not slow him down.  Table of Contents    Answer to 3.

 

4. Phillip is standing on shore attempting to pull a 100. kg box toward him by pulling on a rope. The rope is attached to the box at an angle of 35o to the ice. Assuming no friction between the box and the ice, what is the acceleration of the box toward the shore if he applies a force of 500. N on the rope. What is the normal force of the box on the ice?   Table of Contents   Answer to 4.

 

5. In a Perils of Katie movie Katie(mass = 50.0 kg) and Tony(mass = 65.0 kg) are connected by a cord and are sitting on a frictionless surface. A second cord is attached to Tony and is being pulled horizontally with a force of 600. N toward a bottomless pit. What is the acceleration of Katie and Tony toward the pit? If they start from rest and the pit is 500. m away, how long before they fall into the pit? What is the tension in the rope?   Table of Contents    Answer to 5.

 

6. In a later scene, Katie and Tony are still attached by the rope with the rope hanging over a frictionless, massless pulley(This arrangement is called an Atwood machine). They start of holding onto each other, but they slip out of each others arms. What is the tension on the rope and their accelerations? If the rope is 100. m long and there is a river full of crocodiles 95 m below the pulley, how long before one of them hits the river? Is it Katie or Tony?   Table of Contents    Answer to 6.

 

7. In a still later scene, they have somehow ended up back on the pulley. This time Tony’s assistant,
Brad(mass = 60.0 kg) is hanging on a rope attached to Tony 30.0 m above Tony. How long after Katie slips out of Tony’s arms does Brad have a chance to grab hold of Katie and keep Tony from ending up in the river?  How fast are they moving when Brad grabs for Katie?  Table of Contents     Answer to 7.

  
8. During his Freshman year in college Patrick gets his truck stuck in the sand at the beach. Being a good Physics student, he attaches a rope to the bumper of this truck and to a stout tree in the direction he wants to car to go. He then pushes on the rope sideways at the middle of the rope. When the truck just begins to move, he is pushing at an estimated force of 400. N and the rope is now at an angle of 4.0o. What is the tension in the rope?   Table of Contents    Answer to 8.

    

    

Answers

1. Lauren(mass = 55.0 kg) is laying on a sofa.   Table of Contents
(a)   What is the FN applied by the sofa?
    
FG  =  F N   =  mg = 55.0 kg x 9.80 m s -2  =  539 N    

FG and FN are both acting on Lauren.  We are using Newton's Second Law not the Third Law.
   
(b)		     
What is FN if Allie pushes down on her with a force of 300. N?
   
The total downward force on Lauren is FG + F Allie = 539 N + 300. N = 839 N

FN is the upward force that the sofa must apply to have 0 acceleration.   F N = 839 N.
   
(c)		     
What is FN if Allie is pulling up on her with a force of 300. N?
   
The total downward force on Lauren is FG - F Allie = 539 N - 300. N = 239 N

FN is the upward force that the sofa must apply to have 0 acceleration.   F N = 239 N.
    
(d)		    
What is FN if Allie is pulling up on her with a force of 600. N?
   
Lauren is accelerated upward since the net force on Lauren is 600. N - 539 N = 61 N .

Lauren is no longer in contact with the sofa and FN  = 0 N
    
(e)		    
What is the acceleration for each of the above?
   
While Lauren is in contact with the sofa, a through c, the acceleration is zero.

In part d  

   
            F            61 kg m s -2 
a = ------------ =    ------------------- = 1.1 m s -2  upward
         mass             55.0 kg
    

   
2. Stan(mass = 65.0 kg) is being pulled by Melissa with a force of 600. N and Jamel with a force of 900. N. There is an angle of 35.0o between the two. What happens to Stan?   Table of Contents
   
Find the sum of forces on Stan.  Remember Force is a vector.

Complete the parallelogram.  The resultant vector in red can be calculated using the Law of Cosines.  

The Sum of adjacent angles in a parallelogram is 180o.

c = (a2 + b2 - 2ab Cosine C) 1/2 

c = ( (600. N)2 + (900. N)2 - 2 x 600. N x 900. N x Cosine 145o) 1/2  

c = 1430 N

Now use the Law of Sines to find the Angle to the right of the 900. N force.
   

Sin A      Sin B
-------- = ---------
    a            b 
   

                    a x Sin B                     600. N x Sin 145o 
A = Sin -1 ( -------------- ) = Sin -1 ( --------------------------) = 13.9o  
                          b                                   1430 N
   

There is a 1430 N force 13.9o to the right of the 900. N force applied by Jamel.

F = ma      Solve for a to calculate Stan's Acceleration.  Direction must be the same as the Force.

   
          F         1430 N
a = ----------- = ----------- =  22.0 m s -2  13.9o to the right of Jamel.  
           m           65.0 kg
    

   
3. Tyler(mass = 75.0 kg) is swimming in the ocean. His kick and stroke combine for a forward force of 60.0 N. A current creates a force of 30.0 N and is coming from behind him at an angle 25o to the left. What is his velocity 15.0 seconds after he starts to swim. Assume the water resistance does not slow him down.  Table of Contents
   
A current coming from behind 25o to the left will push Tyler to the right.  

Find the sum of forces on Tyler.  Remember Force is a vector.

Complete the parallelogram.  The resultant vector in red can be calculated using the Law of Cosines.  

The Sum of adjacent angles in a parallelogram is 180o.

c = (a2 + b2 - 2ab Cosine C) 1/2 

c = ( (60.0 N)2 + (30.0 N)2 - 2 x 60.0 N x 30.0 N x Cosine 155o) 1/2  

c = 88.1 N

Now use the Law of Sines to fivnd the Angle to the right of the 60.0 N force.
   

Sin A      Sin B
-------- = ---------
    a            b 
   

                    a x Sin B                     30.0 N x Sin 155o 
A = Sin -1 ( -------------- ) = Sin -1 ( --------------------------) = 8.27o  
                          b                                    88.1 N
   

There is a 88.1 N force 8.27o to the right of the Tylers Force.

F = ma      Solve for a to calculate Tyler's Acceleration.  Direction must be the same as the Force.

   
          F         88.1 N
a = ----------- = ----------- =  1.17 m s -2  8.27o to the right of Tyler's Force.  
           m           75.0 kg

Now solve for the velocity.

v = at = 1.17 m s -2 x 15.0 s = 17.6 m s -1  8.27o to the right of Tyler's Force.  
   

   
4.

 

 Phillip is standing on shore attempting to pull a 100. kg box toward him by pulling on a rope. The rope is attached to the box at an angle of 35o to the ice. Assuming no friction between the box and the ice, what is the acceleration of the box toward the shore if he applies a force of 500. N on the rope. What is the normal force of the box on the ice?   Table of Contents
   
The 500. N force applied by the rope must be broken into horizontal and vertical components.

Vertical Component = 500. N x Sin 35.0o 

                                  = 287 N

  
Horizontal Component = 500. N x Cos 35.0o 

                                                                                                                                = 410. N

Only the Horizontal Component accelerates the box towards the shore.

 

F = ma      Solve for a to calculate the Boxes's Acceleration.  Direction must be the same as the Force.

   
          F        410. N
a = ----------- = ----------- =  4.10 m s -2  toward the shore  
           m           100. kg

  
The Normal Force of the Box on the ice is the same as the Normal Force of the ice on the Box but in opposite directions.

The Total Force of the Box on the ice is mg - Vertical Component of the Rope.

Force Normal = 100. kg x 9.80 m s -2  -  287 N = 693 N
    

   
5.

 

 In a Perils of Katie movie Katie(mass = 50.0 kg) and Tony(mass = 65.0 kg) are connected by a cord and are sitting on a frictionless surface. A second cord is attached to Tony and is being pulled horizontally with a force of 600. N toward a bottomless pit. What is the acceleration of Katie and Tony toward the pit? If they start from rest and the pit is 500. m away, how long before they fall into the pit? What is the tension in the rope?   Table of Contents   
   

 

 

 

 

  

Since it is a frictionless surface with no other forces on the victims, the rope between Katie and Tony has the same tension as the rope attached to just Tony.  If friction is present, the tension will be different.

The 600. N force is applied to a total mass of 115 kg.

 

F = ma      Solve for a to calculate their Acceleration.  Direction must be the same as the Force.

   
          F        600. N
a = ----------- = ----------- =  5.22 m s -2  toward the  bottomless pit
           m           115 kg

The initial velocity is zero.

X = 1/2 at2             Solve for time.

   
           2X                     2 x 500. m
t = ( -------- ) 1/2  =  ( -------------------- ) 1/2  =  13.8 seconds
           a                          5.22 m s -2  
    

    
6.

 

 In a later scene, Katie and Tony are still attached by the rope with the rope hanging over a frictionless, massless pulley(This arrangement is called an Atwood machine). They start of holding onto each other, but they slip out of each others arms. What is the tension on the rope and their accelerations? If the rope is 100. m long and there is a river full of crocodiles 95 m below the pulley, how long before one of them hits the river? Is it Katie or Tony?   Table of Contents
    
The upward force due to tension in the rope is the same for Katie and Tony.  

Write the Sum of Forces = ma separately for each.

Acceleration due to gravity is downward.  Tony with the larger mass will move downward and Katie will move upward.  Set down as the positive direction.
    

mK g - T = - mK a      Katie is moving upward.

mT g - T = mT a         Tony is moving downward.
   

Sove each equation for T.  Then set the two equal to each other.

T = mK g + mK a          T = mT g - mT a

mK g + mK a  = mT g - mT a     Solve for a.

mK a  + mT a  = mT g - mK g 
   

        mT g - mK g 
a = --------------------
           mK  +  mT   

   
         65.0 kg x 9.80 m s -2  -  50.0 kg x 9.80 m s -2 
a = --------------------------------------------------------------- = 1.28 m s -2   downward
                50.0 kg + 65.0 kg 

    
The initial velocity is zero and Tony has to fall 45.0 m before he hits the water.  This is the time when the point where he is tied hits the water.

X = 1/2 at2             Solve for the time before Tony hits the water.

  
           2X                     2 x 45.0. m
t = ( -------- ) 1/2  =  ( -------------------- ) 1/2  =  8.39 seconds  
           a                          1.28 m s -2  
    

    
7.

 

 In a still later scene, they have somehow ended up back on the pulley. This time Tony’s assistant,
Brad(mass = 60.0 kg) is hanging on a rope attached to Tony 30.0 m above Tony. How long after Katie slips out of Tony’s arms does Brad have a chance to grab hold of Katie and keep Tony from ending up in the river?  How fast are they moving when Brad grabs for Katie?   Table of Contents
    
The upward force due to tension in the rope is the same for Katie and Tony.  The tension on the rope between Brad and Tony is different.

Brad and Tony can be treated as a single unit on the left side.
   

Write the Sum of Forces = ma separately for each.

Acceleration due to gravity is downward.  Tony and Brad with the larger mass will move downward and Katie will move upward.  Set down as the positive direction.
    

mK g - T = - mK a        Katie is moving upward.

mT+B  g - T = mT+B a  Tony and Brad are moving downward.
   
   

Sove each equation for T.  Then set the two equal to each other.

T = mK g + mK a          T = mT+B g - mT+B a

mK g + mK a  = mT+B g - mT+B a     Solve for a.

mK a  + mT+B a  = mT+B g - mK g 
    
   
   
    

       mT+B g - mK g         125.0 kg x 9.80 m s -2  - 50.0 kg x 9.80 m s -2 
a = ---------------------- = ------------------------------------------------------------- = 4.20 m s -2  
        mT+B + mK                  125.0 kg + 50.0 kg 

Brad travels 15.0 m down and Katie travels 15.0 m up when they meet.

X = 1/2 at2             Solve for the time when Katie and Brad meet.

  
           2X                     2 x 15.0. m
t = ( -------- ) 1/2  =  ( -------------------- ) 1/2  =  2.67 seconds  
           a                          4.20 m s -2  

    
v = at  = 4.20 m s -2 x 2.67 s = 11.2 m s -1       

Brad is moving 11.2 m s -1 downward at the same time Katie is moving upward at 11.2 m s -1 .
    

    
8. During his Freshman year in college Patrick gets his truck stuck in the sand at the beach. Being a good Physics student, he attaches a rope to the bumper of this truck and to a stout tree in the direction he wants to car to go. He then pushes on the rope sideways at the middle of the rope. When the truck just begins to move, he is pushing at an estimated force of 400. N and the rope is now at an angle of 4.0o. What is the tension in the rope?   Table of Contents
  
The force on the rope is the hypotenuse of the Right Triangle formed.

Solve Sine of a Right Triangle for the hypotenuse. 

  
              opposite
Sine = ----------------
            hypotenuse
   

                       opposite
hypotenuse = ------------
                         Sine
  

                        400. N
hypotenuse = -----------
                         Sin 4.0o
   

hypotenuse = 5730 N