A spherical ball with a mass of 2.00 kg rests in the notch
shown below. If there is no friction between the ball and
the walls, what is the magnitude of the force exerted on
the ball by wall1?
Answer: [XXXXXXXXXXXXXXXX]
Id  Name  Action  Entries 

(PROJECTION (COMPO X 0 (FORCE BALL WALL1 NORMAL :TIME 1)))  projection equations  writing the projection equation for the normal force on the ball at T0 due to wall1 onto the x axis 

(PROJECTION (COMPO X 0 (FORCE BALL EARTH WEIGHT :TIME 1)))  projection equations  writing the projection equation for the weight force on the ball at T0 due to the earth onto the x axis 

(COMPOEQN NFL X 0 (NL BALL 1))  Newton's second law  applying Newton's second law to the ball at T0 

(PROJECTION (COMPO X 0 (FORCE BALL WALL2 NORMAL :TIME 1)))  projection equations  writing the projection equation for the normal force on the ball at T0 due to wall2 onto the x axis 

(PROJECTION (COMPO Y 0 (FORCE BALL WALL2 NORMAL :TIME 1)))  projection equations  writing the projection equation for the normal force on the ball at T0 due to wall2 onto the y axis 

(STDCONSTANT (GRAVITATIONALACCELERATION EARTH))  value of g on Earth  defining the value of g on Earth  
(GIVEN (MASS BALL) (DNUM 2.0 kg))  the given value  entering the given value of the mass of the ball  
(WTLAW BALL 1)  Weight law  applying the Weight law on the ball 

(PROJECTION (COMPO Y 0 (FORCE BALL EARTH WEIGHT :TIME 1)))  projection equations  writing the projection equation for the weight force on the ball at T0 due to the earth onto the y axis 

(COMPOEQN NFL Y 0 (NL BALL 1))  Newton's second law  applying Newton's second law to the ball at T0 

(PROJECTION (COMPO Y 0 (FORCE BALL WALL1 NORMAL :TIME 1)))  projection equations  writing the projection equation for the normal force on the ball at T0 due to wall1 onto the y axis 

Id  Name  Symbol  Entries 

(COMPO Y 0 (FORCE BALL WALL1 NORMAL :TIME 1))  the y component of the normal force on the ball at T0 due to wall1  Yc_Fn_BALL_WALL1_1_0  
(GRAVITATIONALACCELERATION EARTH)  the gravitational acceleration due to the earth  g_EARTH  
(MASS BALL)  the mass of the ball  m_BALL  
(MAG (FORCE BALL EARTH WEIGHT :TIME 1))  the magnitude of the weight force on the ball at T0 due to the earth  Fw_BALL_EARTH_1  
(COMPO Y 0 (FORCE BALL EARTH WEIGHT :TIME 1))  the y component of the weight force on the ball at T0 due to the earth  Yc_Fw_BALL_EARTH_1_0  
(COMPO Y 0 (FORCE BALL WALL2 NORMAL :TIME 1))  the y component of the normal force on the ball at T0 due to wall2  Yc_Fn_BALL_WALL2_1_0  
(MAG (FORCE BALL WALL2 NORMAL :TIME 1))  the magnitude of the normal force on the ball at T0 due to wall2  Fn_BALL_WALL2_1  
(COMPO X 0 (FORCE BALL WALL2 NORMAL :TIME 1))  the x component of the normal force on the ball at T0 due to wall2  Xc_Fn_BALL_WALL2_1_0  
(COMPO X 0 (FORCE BALL EARTH WEIGHT :TIME 1))  the x component of the weight force on the ball at T0 due to the earth  Xc_Fw_BALL_EARTH_1_0  
(COMPO X 0 (FORCE BALL WALL1 NORMAL :TIME 1))  the x component of the normal force on the ball at T0 due to wall1  Xc_Fn_BALL_WALL1_1_0  
(MAG (FORCE BALL WALL1 NORMAL :TIME 1))  the magnitude of the normal force on the ball at T0 due to wall1  Fn_BALL_WALL1_1 
Entry  Operator  Hints  

(BODY BALL) 
(DRAWBODY BALL NIL)  It is a good idea to begin by choosing the body or system of bodies you are going to focus on.  …  You should use the body tool to draw a body choosing the ball as the body. 
(DEFINEVAR (GRAVITATIONALACCELERATION EARTH)) 
(DEFINEGRAVACCEL EARTH)  Define a variable for the gravitational acceleration due to the earth by using the Add Variable command on the Variable menu and selecting gravitational acceleration.  
(DEFINEVAR (MASS BALL)) 
(DEFINEMASS BALL NIL)  You can use the variable definition tools, which are under the variables menu, in order to define a variable for mass.  
(DRAWAXES 0) 
(DRAWUNROTATEDAXES)  …  Draw a standard horizontalvertical coordinate system setting the positive x axis at 0 degrees.  
(DRAWVECTORALIGNEDAXES 0)  What would be a useful choice for the coordinate axes?  …  Rotate the coordinate axes setting the x axis at 0 degrees.  
(DRAWPROJECTIONAXES 0)  
(EQN (= Fw_BALL_EARTH_1 (* m_BALL g_EARTH))) 
(WTLAW BALL 1)  Try applying the weight law.  …  Write Fw_BALL_EARTH_1 = m_BALL*g_EARTH 
(EQN (= Xc_Fn_BALL_WALL1_1_0 (* Fn_BALL_WALL1_1 (COS ( (DNUM 120 deg) (DNUM 0 deg)))))) 
(COMPOGENERALCASE X 0 (FORCE BALL WALL1 NORMAL :TIME 1))  Since the normal force on the ball at T0 due to wall1 at T0 is not perpendicular to the NIL axis, it has a nonzero component along that axis.  …  Since the direction of the normal force on the ball at T0 due to wall1 at T0 is $qFn_BALL_WALL1_1 (120 deg) and the orientation of the x axis is $qNIL (0 deg), you can write the general formula Xc_Fn_BALL_WALL1_1_0 = Fn_BALL_WALL1_1 * cos($qFn_BALL_WALL1_1  $qNIL). 
(EQN (= Xc_Fn_BALL_WALL2_1_0 (* Fn_BALL_WALL2_1 (COS ( (DNUM 40 deg) (DNUM 0 deg)))))) 
(COMPOGENERALCASE X 0 (FORCE BALL WALL2 NORMAL :TIME 1))  Since the normal force on the ball at T0 due to wall2 at T0 is not perpendicular to the NIL axis, it has a nonzero component along that axis.  …  Since the direction of the normal force on the ball at T0 due to wall2 at T0 is $qFn_BALL_WALL2_1 (40 deg) and the orientation of the x axis is $qNIL (0 deg), you can write the general formula Xc_Fn_BALL_WALL2_1_0 = Fn_BALL_WALL2_1 * cos($qFn_BALL_WALL2_1  $qNIL). 
(EQN (= Xc_Fw_BALL_EARTH_1_0 0)) 
(COMPOPERPENDICULAR X 0 (FORCE BALL EARTH WEIGHT :TIME 1))  Notice that at T0, the weight force on the ball at T0 due to the earth is perpendicular to the NIL axis.  …  Because the weight force on the ball at T0 due to the earth is perpendicular to the NIL axis at T0, write the equation Xc_Fw_BALL_EARTH_1_0=0 
(EQN (= Yc_Fn_BALL_WALL1_1_0 (* Fn_BALL_WALL1_1 (SIN ( (DNUM 120 deg) (DNUM 0 deg)))))) 
(COMPOGENERALCASE Y 0 (FORCE BALL WALL1 NORMAL :TIME 1))  Since the normal force on the ball at T0 due to wall1 at T0 is not perpendicular to the NIL axis, it has a nonzero component along that axis.  …  Since the direction of the normal force on the ball at T0 due to wall1 at T0 is $qFn_BALL_WALL1_1 (120 deg) and the orientation of the x axis is $qNIL (0 deg), you can write the general formula Yc_Fn_BALL_WALL1_1_0 = Fn_BALL_WALL1_1 * sin($qFn_BALL_WALL1_1  $qNIL). 
(EQN (= Yc_Fn_BALL_WALL2_1_0 (* Fn_BALL_WALL2_1 (SIN ( (DNUM 40 deg) (DNUM 0 deg)))))) 
(COMPOGENERALCASE Y 0 (FORCE BALL WALL2 NORMAL :TIME 1))  Since the normal force on the ball at T0 due to wall2 at T0 is not perpendicular to the NIL axis, it has a nonzero component along that axis.  …  Since the direction of the normal force on the ball at T0 due to wall2 at T0 is $qFn_BALL_WALL2_1 (40 deg) and the orientation of the x axis is $qNIL (0 deg), you can write the general formula Yc_Fn_BALL_WALL2_1_0 = Fn_BALL_WALL2_1 * sin($qFn_BALL_WALL2_1  $qNIL). 
(EQN (= Yc_Fw_BALL_EARTH_1_0 ( Fw_BALL_EARTH_1))) 
(COMPOPARALLELAXIS Yc_Fw_BALL_EARTH_1_0)  Since the weight force on the ball at T0 due to the earth at T0 lies along the NIL axis, it has a nonzero component along that axis.  …  Since the weight force on the ball at T0 due to the earth at T0 lies along the y axis and points in the  y direction, write the equation Yc_Fw_BALL_EARTH_1_0 =  Fw_BALL_EARTH_1. 
(EQN (= g_EARTH (DNUM 9.8 m/s^2))) 
(WRITEGONEARTH)  You should know the value of g for the Earth  …  Write the equation g_EARTH = 9.8 m/s^2 
(EQN (= m_BALL (DNUM 2.0 kg))) 
(WRITEKNOWNVALUEEQN (MASS BALL))  You can find the value of the mass of the ball in the problem statement.  The value of the mass of the ball is given as 2.0 kg.  Enter the equation m_BALL = 2.0 kg. 
(EQN (= (+ Xc_Fw_BALL_EARTH_1_0 Xc_Fn_BALL_WALL2_1_0 Xc_Fn_BALL_WALL1_1_0) 0)) 
(WRITENFLCOMPO BALL 1 X 0)  You can apply Newton's second law to the ball. Note that the ball is not accelerating at T0.  …  Because the ball is not accelerating at T0, write Newton's second law as Xc_Fw_BALL_EARTH_1_0 + Xc_Fn_BALL_WALL2_1_0 + Xc_Fn_BALL_WALL1_1_0 = 0 
(EQN (= (+ Yc_Fw_BALL_EARTH_1_0 Yc_Fn_BALL_WALL2_1_0 Yc_Fn_BALL_WALL1_1_0) 0)) 
(WRITENFLCOMPO BALL 1 Y 0)  You can apply Newton's second law to the ball. Note that the ball is not accelerating at T0.  …  Because the ball is not accelerating at T0, write Newton's second law as Yc_Fw_BALL_EARTH_1_0 + Yc_Fn_BALL_WALL2_1_0 + Yc_Fn_BALL_WALL1_1_0 = 0 
(IMPLICITEQN (= OFn_BALL_WALL1_1 (DNUM 120 deg))) 
(WRITEIMPLICITEQN (DIR (FORCE BALL WALL1 NORMAL :TIME 1)))  
(IMPLICITEQN (= OFn_BALL_WALL2_1 (DNUM 40 deg))) 
(WRITEIMPLICITEQN (DIR (FORCE BALL WALL2 NORMAL :TIME 1)))  
(IMPLICITEQN (= OFw_BALL_EARTH_1 (DNUM 270 deg))) 
(WRITEIMPLICITEQN (DIR (FORCE BALL EARTH WEIGHT :TIME 1)))  
(IMPLICITEQN (= Xc_Fw_BALL_EARTH_1_0 0)) 
(COMPOPARALLELAXIS Yc_Fw_BALL_EARTH_1_0)  Since the weight force on the ball at T0 due to the earth at T0 lies along the NIL axis, it has a nonzero component along that axis.  …  Since the weight force on the ball at T0 due to the earth at T0 lies along the y axis and points in the  y direction, write the equation Yc_Fw_BALL_EARTH_1_0 =  Fw_BALL_EARTH_1. 
(IMPLICITEQN (= Xc_a_BALL_1_0 0)) 
(WRITENFLCOMPO BALL 1 X 0)  You can apply Newton's second law to the ball. Note that the ball is not accelerating at T0.  …  Because the ball is not accelerating at T0, write Newton's second law as Xc_Fw_BALL_EARTH_1_0 + Xc_Fn_BALL_WALL2_1_0 + Xc_Fn_BALL_WALL1_1_0 = 0 
(IMPLICITEQN (= Yc_a_BALL_1_0 0)) 
(WRITENFLCOMPO BALL 1 Y 0)  You can apply Newton's second law to the ball. Note that the ball is not accelerating at T0.  …  Because the ball is not accelerating at T0, write Newton's second law as Yc_Fw_BALL_EARTH_1_0 + Yc_Fn_BALL_WALL2_1_0 + Yc_Fn_BALL_WALL1_1_0 = 0 
(IMPLICITEQN (= a_BALL_1 (DNUM 0 m/s^2))) 
(WRITEIMPLICITEQN (MAG (ACCEL BALL :TIME 1)))  
(VECTOR (ACCEL BALL :TIME 1) ZERO) 
(ACCELATREST BALL 1)  Notice that the ball is at rest at T0.  …  Because the ball is at rest at T0, use the acceleration tool to draw a zerolength acceleration vector for it. 
(VECTOR (FORCE BALL EARTH WEIGHT :TIME 1) (DNUM 270 deg)) 
(DRAWWEIGHT BALL 1 EARTH)  Notice that the ball is near the earth.  …  Because the ball is near the planet the earth, the planet exerts a weight force on it, so use the force drawing tool to draw a force on the ball due to the earth of type weight at T0. 
(VECTOR (FORCE BALL WALL1 NORMAL :TIME 1) (DNUM 120 deg)) 
(DRAWNORMAL BALL WALL1 1)  Notice that the ball is supported by a surface: wall1.  …  Because wall1 supports the ball, draw a normal force on the ball due to wall1 at an angle of 120 degrees degrees. 
(VECTOR (FORCE BALL WALL2 NORMAL :TIME 1) (DNUM 40 deg)) 
(DRAWNORMAL BALL WALL2 1)  Notice that the ball is supported by a surface: wall2.  …  Because wall2 supports the ball, draw a normal force on the ball due to wall2 at an angle of 40 degrees degrees. 