(a)  What is the force applied to the leg?

(b)  What holds the person in place in the bed?

 Answer to #1

Calculate if the cable to the left exerts 75.0 N and the cable up and to the right exerts 550. N.

Calculate the force on the cable if the cable dropped straight down from the ceiling and the mass of the chandelier.

Why would we use the arrangement shown to the left instead of having the cable drop straight down from the ceiling?

Answer to #3   

The see-saw above has a mass of 50.0 kg that has its center of gravity above the fulcrum.  Ralph, 72.0 kg, is on the left 225 cm from the fulcrum.  Cindy, 56.5 kg, is on the right.  (a) How far from the fulcrum must Cindy sit to be in equilibrium?  (b) What is the force exerted by the fulcrum on the see-saw?   Answer to #4

 Answer to #5  

 Answer to #6  

 Answer to #7  

(a)  Calculate the force of the ladder on the wall.

(b)  Calculate the force of the ladder on the ground.

   Answer to #8

(a)  Calculate the force on the cable attached to the end of the beam and the top of the vertical support.

(b)  Calculate the forces where the beam is attached to the support.

 Answer to #9  

The traffic signal to the left has a mass of 85.5 kg.  If is 15.0^o, what is the force on each cable? 

  Answer to #10 

A 65.6 kg traffic light is attached to the end of a 125 kg uniform beam that is 3.00 m long.  The beam is attached to a vertical support with a of 40.0^o and 1.50 m from the top of the vertical support.  A cable is attached at a right angle from the top of the vertical support. 

(a)  Calculate d₂, the distance from where the cable attaches to the uniform beam to the end of the beam.

(b)  The force on the cable.

(c)  The forces where the beam is attached to the vertical support.

  Answer to #11

(a)  What is the force applied to the leg?

(b)  What holds the person in place in the bed?

This is the tension on the rope both on the top and the bottom.  Tension is always a pull.  The vertical components cancel each other.  Only the horizontal component is stretching the leg.

F_horizontal  =   2 x Tension x Cos 35.0^o  =  2 x 245 N x Cos 35.0^o  =  401 N

Observe that the horizontal force on the leg is greater than the force created by the mass.  If the person tries to raise or lower his leg, it will create an imbalance in the vertical components because the angles will no longer be equal.  This will accelerate the leg in the direction to put in back in the proper allignment.

The person is held in place in the bed by friction.  If the desired force on the leg is greater than friction can create, the person would have to be strapped into the bed.
   

Total downward force = 141.0 kg x 9.80 m s ^-2  =  1380 N

Downward force on scale = 15.0 kg x 9.80 m s ^-2  =  147 N

   
Force Exerted = Total downward force  -  Downward force on scale  =  1380 N  -  147 N  =  1230 N
   

Calculate if the cable to the left exerts 75.0 N and the cable up and to the right exerts 550. N.

Calculate the force on the cable if the cable dropped straight down from the ceiling and the mass of the chandelier.

Why would we use the arrangement shown to the left instead of having the cable drop straight down from the ceiling?

   
Force_right  =  Force_left  =  75.0 N            This is the horizontal component at .  
   

                           F _horizontal                         75.0 N
      =  Cos ^-1 ( ----------------- )  =  Cos ^-1 ( ----------- ) =  82.2^o  
                           F _tension                              550. N

     =  550. N x Sin 82.2^o  =  545 N

    The chandelier only has a vertical component.  The vertical component of
    the cable must match this force in the opposite direction.

                 
   
               F              545 N
mass =  -----  =  -----------------  =  55.6 kg
               g           9.80 m s ^-2  
   

The cable must be attached to a support strong enough for the applied force.  If a support is not in the proper location, this can be used to get the chandelier in the desired location without adding additional supports.
    

The see-saw above has a mass of 50.0 kg that has its center of gravity above the fulcrum.  Ralph, 72.0 kg, is on the left 225 cm from the fulcrum.  Cindy, 56.5 kg, is on the right.  (a) How far from the fulcrum must Cindy sit to be in equilibrium?  (b) What is the force exerted by the fulcrum on the see-saw?

  F _Ralph  =  72.0 kg x 9.80 m s ^-2  =  706 N

  F _Cindy  =  56.5 kg x 9.80 m s ^-2  =  554 N

  F _board  =  50.0 kg x 9.80 m s ^-2  =  490. N

Counter-clockwise torque  =   Clockwise torque        Use the fulcrum as the axis of rotation.  Since the center of gravity is at the fulcrum, the torque at this point is zero.

F _Ralph d _Ralph  =  F _Cindy d _Cindy   
   

                  F _Ralph d _Ralph         706 N x 225 cm
d _Cindy  =  --------------------  =  -----------------------  =  287 cm from the fulcurm
                      F _Cindy                       554 N
   

All of the above forces are downward, the only upward force at the fulcrum must balance the total of these forces.

F _N  =  706 N  +  554 N  +  490. N  =  1750. N
   

   
F = 0   and  = 0

d₁ = 1.50 m           d₂  =  0.25 m           d₃  =  1.25 m

F  =  mg

F _R  =  350. kg  x  9.80 m s ^-2  =  3430 N

F _CG  =  555 kg x 8.90 m s ^-2  =  5440 N

Use F₁ as the axis of rotation.  This leaves only one unknown term.

Clockwise Torque = Counter-clockwise Torque

F_CG d₁  +  F_R (d₁ + d₂)  =  F₂ (d₁ + d₂ + d₃)                        Solve for F₂.
   

            F_CG d₁  +  F_R (d₁ + d₂)          5440 N x 1.50 m  +  3430 N x 1.75 m
F₂  =  ---------------------------------  = ----------------------------------------------------  =  4720 N
               (d₁ + d₂ + d₃)                                     3.00 m

Since all the forces except one are now known, we can use Sum of Forces to solve for the unknown.  Set up forces as positive and down forces as negative.

F₁  +  F₂  -  F_R  -  F_CG  =  0                            Solve for F₁.

F₁  =  -  F₂  +  F_R  +  F_CG  =  - 4720 N  +  5440 N  +  3430 N  =  4150 N
    

   
F = 0   and  = 0

d₁  =  1.50 m        d₂  =  3.50 m           d_CG  =  2.50 m      (d_CG = distance from either end to F_CG)

F  =  mg

F _CG  =  100. kg  x  9.80 m s ^-2  =  980. N

Use F₁ as the axis of rotation.  This leaves only one unknown term.

Clockwise Torque = Counter-clockwise Torque

F₂d₁  =  F_CG d_CG                            Solve for F₂.
   

            F_CG d_CG         980. N x 2.50 m
F₂  =  --------------  =  -----------------------  =  1630 N
               d₁                     1.50 m
   

Since all the forces except one are now known, we can use Sum of Forces to solve for the unknown.  Set up forces as positive and down forces as negative.

F₁  +  F₂  -  F_CG  =  0                                     Solve for F₁.

F₁  =  F_CG  -  F₂  =  980. N  -  1630 N  =  -650 N

Although we drew F₁ as an upward force, it is actually and downward force.   This affects the type of connection necessary when building the cantilever.
   

  
F = 0   and  = 0

d₁  =  2.00 m        d₂  =  2.00 m       d₃ = 4.00 cm    d_CG  =  3.00 m      (d_CG = distance from either end to F_CG)

F  =  mg

F _CG  =  500. kg  x  9.80 m s ^-2  =  4.90 x 10³ N

F_safe  =  1000. kg x 9.80 m s ^-2  =  9.80 x 10³ N

Use F₁ as the axis of rotation.  This leaves only one unknown term.

Clockwise Torque = Counter-clockwise Torque

F₂d₁  =  F_CG d_CG  + F_safed_safe                         Solve for F₂.

            F_CG d_CG    + F_safed_safe         4.90 x 10³ N x 3.00 m  +  9.80 x 10³ N x 4.00 m
F₂  =  --------------------------------  =  -----------------------------------------------------------------  =  2.70 x 10⁴ N
                              d₁                                                      2.00 m
   

Since all the forces except one are now known, we can use Sum of Forces to solve for the unknown.  Set up forces as positive and down forces as negative.

F₁  +  F₂  -  F_CG  - F_safe  =  0                                     Solve for F₁.

F₁  =  F_CG  +  F_safe  -  F₂  =  4.90 x 10³ N  +  9.80 x 10³ N  -  2.70 x 10⁴ N  =  - 1.23 x 10⁴ N

Although we drew F₁ as an upward force, it is actually and downward force.   This affects the type of connection necessary when building the cantilever.
   

(a)  Calculate the force of the ladder on the wall.

(b)  Calculate the force of the ladder on the ground.

F = 0   and  = 0

d₁  =  1.00 m        d₂  =  5.50 m       

d_ladder = ( ( 1.00 m )²  +  ( 5.50 m )² ) ^1/2  =  5.59 m
    

                      d₂                         5.50 m
  = Tan ^-1 ( ----- )  =  Tan ^-1 ( ------------ ) = 79.7^o
                      d₁                        1.00 m
   

d_CG  =  2.80 m      (d_CG = distance from either end to F_CG)

F  =  mg

F _CG  =  45.0 kg  x  9.80 m s ^-2  =  441 N

Use where ladder touches the ground as the axis of rotation.  This leaves only one unknown term.  All forces are forces on the ladder.  The ladder applies an equal but opposite force on both the wall and the grass.

Clockwise Torque = Counter-clockwise Torque

d_cgF_CG  x Cos  =  d_ladder x F _{N ladder}                   Solve for F _{N ladder} 

   
                     d_cgF_CG  x Cos           2.80 m x 441 N x Cos 79.7^o
F_{N ladder}  =  -----------------------  =  --------------------------------------  =  221 N
                           d_ladder                                5.59 m
   

F _{N Wall}  =  F _{N ladder} x Cos  =  221 N x Cos 79.7^o  =  39.5 N

F _{P Wall}  =  F _{N ladder} x Sin  =  221 N x Sin 79.7^o  =  217 N

   
F _Horizontal   = 0   and  F _Vertical   = 0              These are the forces on the ladder.  
   

F _down  =  F _up       F _{W ladder}   =  F _{N Grass}  +  F _{P Wall}           Solve for F _{N Grass.}
   
F _{N Grass}  =  F _{W ladder}  -   F _{P Wall}  =  441 N  -  217 N  =  224 N
   

F _Left  =  F _Right        F _{N Wall}  =  F _{P Grass}  =  39.5 N
   

The Ladder applies an equal and opposite force on the Grass.

F _{Ladder on Grass}  =  ( F _Left²  +  F _Down² ) ^1/2 

F _{Ladder on Grass}   =  ( ( 39.5 N )²  +  ( 224 N )² ) ^1/2  

F _{Ladder on Grass}  =  227 N

    
                       F _Left                            39.5N
A = Tan ^{- 1} ( ----------- )  =  Tan ^-1 ( ------------ )  =  10.0^o
                      F _Down                           224 N
   

(a)  Calculate the force on the cable attached to the end of the beam and the top of the vertical support.

(b)  Calculate the forces where the beam is attached to the support.

F = 0   and  = 0

d₁  =  1.75 m        d₂  =  2.00 m    d₃  =  1.00 m

d _CG  =  1.00 m   from either end

F = mg

F _sign  =  500. kg x 9.80 m s ^-2  =  4.90 x 10 ³ N

F _CG =  250. kg x  9.80 m s ^-2  =  2450 N
   

                       d₃                     1.00 m
    = Tan ^-1 ( ---- ) = Tan ^-1 ( ----------- ) = 26.6^o
                       d₂                     2.00 m
    

Use where beam touches the support as the axis of rotation.  This leaves only one unknown term.  The forces are forces on the beam.  The beam applies an equal but opposite force on support and cable.

Clockwise Torque  =  Counter-clockwise Torque

F _CG  x  d _CG  +  F _sign x d₁  =  F _cable x d₂ x Sin                 Solve for F _cable  
    

                  F _CG  x  d _CG  +  F _sign x d₁           2450 N x 1.00 m  +  4.90 x 10 ³ N x 1.75 m
F _cable  =  -------------------------------------  =  ------------------------------------------------------------  =   1.23 x 10 ⁴ N
                            d₂ x Sin                                           2.00 m x Sin 26.6^o    
   

F _{up cable}  =  F _cable x Sin   =  1.23 x 10 ⁴ N  x  Sin 26.6^o  =  5510 N

F _{left cable}  =  F _cable x Cos   =  1.23 x 10 ⁴ N  x  Cos 26.6^o  = 1.10 x 10 ⁴ N

The Support applies equal forces in opposite directions on the cable.

   
F _Horizontal   = 0   and  F _Vertical   = 0              These are the forces on the beam.  
   

F _down  =  F _up       F _CG  +  F _sign   =  F _{up cable}  +  F _{up Support}           Solve for F _{up Support.}
   
F _{up Support}  =  F _CG  +   F _sign  - F _{up cable}  =  2450 N  +  4.90 x 10 ³ N  -  5510 N  =  1840 N
   

F _Left  =  F _Right        F _{left cablel}  =  F _{right support}  =  1.10 x 10 ⁴ N
   

The Beam applies an equal and opposite force on the Support.

F _{beam on support}  =  ( F _Left²  +  F _Down² ) ^1/2 

F _{beam on support}   =  ( ( 1.10 x 10 ⁴ N )²  +  ( 1840 N )² ) ^1/2  

F _{beam on support}  =  2140 N

    
                       F _Left                            1.10 x 10 ⁴ N
A = Tan ^{- 1} ( ----------- )  =  Tan ^-1 ( ------------------- )  =  30.9^o
                      F _Down                              1840 N
   

F = 0   and  = 0

  =  15.0^o  

F = mg

F _sign  =  85.5 kg x 9.80 m s ^-2  =  838 N

F _up  =  F _down  

2 x F _cable Sin  =  F _sign         Solve for F _cable .

   
                   F _sign                       838 N
F _cable  =  ------------------  =  ---------------------  =  1620 N
                 2 x Sin                 2 x Sin 15.0^o     
   

A 65.6 kg traffic light is attached to the end of a 125 kg uniform beam that is 3.00 m long.  The beam is attached to a vertical support with a of 40.0^o and 1.50 m from the top of the vertical support.  A cable is attached at a right angle from the top of the vertical support. 

(a)  Calculate d₂, the distance from where the cable attaches to the uniform beam to the end of the beam.

(b)  The force on the cable.

(c)  The forces where the beam is attached to the vertical support.

F = 0   and  = 0

  =  40.0^o  

F = mg

F _sign  =  65.6 kg x 9.80 m s ^-2  

F _sign  =  643 N
   

F _CG  =  125 kg x 9.80 m s ^-2  

F _CG  =  1230 N

d ₁  =  1.50 m

length _beam  =  3.00 m

                               d₁             1.50 m
d _{beam to cable}  =  ---------  =  ----------------  =  1.96 m
                            Cos          Cos 40.0^o  

   
Use where beam touches the support as the axis of rotation.  This leaves only one unknown term.  The forces are forces on the beam.  The beam applies an equal but opposite force on support and cable.

Clockwise Torque  =  Counter-clockwise Torque

length _beam x F_sign Sin   +  length _beam x F_CG Sin   =  d _{beam to cable}  x F _cable         Solve for F _cable .

    
                  length _beam x F_sign Sin   +  length _beam x F_CG Sin   
F _cable  =  -------------------------------------------------------------------------
                                     d _{beam to cable}   
    

                        3.00 m x 643 N x Sin 40.0^o  + x 3.00 m x 1230 N x Sin 40.0^o 
F _cable  =  -------------------------------------------------------------------------------------  =  1240 N
                                             1.96 m 
    

F _up  =  F _down         F _{beam up}  =  F _CG  +  F _sign   =  1230 N  +  643 N  =  1870 N
   

F _right  =  F _left        F _{beam right}  =  F _cable  =  1240 N
    

Beam applies an opposite and equal force on the Support.

F _{beam on support}  =  ( F _Left²  +  F _Down² ) ^1/2 

F _{beam on support}   =  ( ( 1240 N )²  +  ( 1870 N )² ) ^1/2  

F _{beam on support}  =  2240 N

    
                       F _Down                         1870 N
A = Tan ^{- 1} ( ----------- )  =  Tan ^-1 ( ------------- )  =  56.5^o
                      F _Left                            1240 N