Problem Solutions for S2E

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]

Solution 3 (out of 6)

Principles (Problem Solving Methods)
IdNameActionEntries
(PROJECTION (COMPO X 0 (FORCE BALL WALL1 NORMAL :TIME 1)))projection equationswriting the projection equation for the normal force on the ball at T0 due to wall1 onto the x axis

(DRAW-AXES 0)

(EQN (= |Xc_Fn_BALL_WALL1_1_0| (* |Fn_BALL_WALL1_1| (COS (- (DNUM 120 |deg|) (DNUM 0 |deg|))))))

(IMPLICIT-EQN (= |OFn_BALL_WALL1_1| (DNUM 120 |deg|)))

(VECTOR (FORCE BALL WALL1 NORMAL :TIME 1) (DNUM 120 |deg|))

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

(DRAW-AXES 0)

(EQN (= |Xc_Fw_BALL_EARTH_1_0| 0))

(IMPLICIT-EQN (= |OFw_BALL_EARTH_1| (DNUM 270 |deg|)))

(VECTOR (FORCE BALL EARTH WEIGHT :TIME 1) (DNUM 270 |deg|))

(COMPO-EQN NFL X 0 (NL BALL 1))Newton's second lawapplying Newton's second law to the ball at T0

(BODY BALL)

(DRAW-AXES 0)

(EQN (= (+ |Xc_Fw_BALL_EARTH_1_0| |Xc_Fn_BALL_WALL2_1_0| |Xc_Fn_BALL_WALL1_1_0|) 0))

(IMPLICIT-EQN (= |OFn_BALL_WALL1_1| (DNUM 120 |deg|)))

(IMPLICIT-EQN (= |OFn_BALL_WALL2_1| (DNUM 40 |deg|)))

(IMPLICIT-EQN (= |OFw_BALL_EARTH_1| (DNUM 270 |deg|)))

(IMPLICIT-EQN (= |Xc_a_BALL_1_0| 0))

(IMPLICIT-EQN (= |a_BALL_1| (DNUM 0 |m/s^2|)))

(VECTOR (ACCEL BALL :TIME 1) ZERO)

(VECTOR (FORCE BALL EARTH WEIGHT :TIME 1) (DNUM 270 |deg|))

(VECTOR (FORCE BALL WALL1 NORMAL :TIME 1) (DNUM 120 |deg|))

(VECTOR (FORCE BALL WALL2 NORMAL :TIME 1) (DNUM 40 |deg|))

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

(DRAW-AXES 0)

(EQN (= |Xc_Fn_BALL_WALL2_1_0| (* |Fn_BALL_WALL2_1| (COS (- (DNUM 40 |deg|) (DNUM 0 |deg|))))))

(IMPLICIT-EQN (= |OFn_BALL_WALL2_1| (DNUM 40 |deg|)))

(VECTOR (FORCE BALL WALL2 NORMAL :TIME 1) (DNUM 40 |deg|))

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

(DRAW-AXES 0)

(EQN (= |Yc_Fn_BALL_WALL2_1_0| (* |Fn_BALL_WALL2_1| (SIN (- (DNUM 40 |deg|) (DNUM 0 |deg|))))))

(IMPLICIT-EQN (= |OFn_BALL_WALL2_1| (DNUM 40 |deg|)))

(VECTOR (FORCE BALL WALL2 NORMAL :TIME 1) (DNUM 40 |deg|))

(STD-CONSTANT (GRAVITATIONAL-ACCELERATION EARTH))value of g on Earthdefining the value of g on Earth

(DEFINE-VAR (GRAVITATIONAL-ACCELERATION EARTH))

(EQN (= |g_EARTH| (DNUM 9.8 |m/s^2|)))

(GIVEN (MASS BALL) (DNUM 2.0 |kg|))the given valueentering the given value of the mass of the ball

(DEFINE-VAR (MASS BALL))

(EQN (= |m_BALL| (DNUM 2.0 |kg|)))

(WT-LAW BALL 1)Weight lawapplying the Weight law on the ball

(DEFINE-VAR (GRAVITATIONAL-ACCELERATION EARTH))

(DEFINE-VAR (MASS BALL))

(EQN (= |Fw_BALL_EARTH_1| (* |m_BALL| |g_EARTH|)))

(IMPLICIT-EQN (= |OFw_BALL_EARTH_1| (DNUM 270 |deg|)))

(VECTOR (FORCE BALL EARTH WEIGHT :TIME 1) (DNUM 270 |deg|))

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

(DRAW-AXES 0)

(EQN (= |Yc_Fw_BALL_EARTH_1_0| (- |Fw_BALL_EARTH_1|)))

(IMPLICIT-EQN (= |OFw_BALL_EARTH_1| (DNUM 270 |deg|)))

(IMPLICIT-EQN (= |Xc_Fw_BALL_EARTH_1_0| 0))

(VECTOR (FORCE BALL EARTH WEIGHT :TIME 1) (DNUM 270 |deg|))

(COMPO-EQN NFL Y 0 (NL BALL 1))Newton's second lawapplying Newton's second law to the ball at T0

(BODY BALL)

(DRAW-AXES 0)

(EQN (= (+ |Yc_Fw_BALL_EARTH_1_0| |Yc_Fn_BALL_WALL2_1_0| |Yc_Fn_BALL_WALL1_1_0|) 0))

(IMPLICIT-EQN (= |OFn_BALL_WALL1_1| (DNUM 120 |deg|)))

(IMPLICIT-EQN (= |OFn_BALL_WALL2_1| (DNUM 40 |deg|)))

(IMPLICIT-EQN (= |OFw_BALL_EARTH_1| (DNUM 270 |deg|)))

(IMPLICIT-EQN (= |Yc_a_BALL_1_0| 0))

(IMPLICIT-EQN (= |a_BALL_1| (DNUM 0 |m/s^2|)))

(VECTOR (ACCEL BALL :TIME 1) ZERO)

(VECTOR (FORCE BALL EARTH WEIGHT :TIME 1) (DNUM 270 |deg|))

(VECTOR (FORCE BALL WALL1 NORMAL :TIME 1) (DNUM 120 |deg|))

(VECTOR (FORCE BALL WALL2 NORMAL :TIME 1) (DNUM 40 |deg|))

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

(DRAW-AXES 0)

(EQN (= |Yc_Fn_BALL_WALL1_1_0| (* |Fn_BALL_WALL1_1| (SIN (- (DNUM 120 |deg|) (DNUM 0 |deg|))))))

(IMPLICIT-EQN (= |OFn_BALL_WALL1_1| (DNUM 120 |deg|)))

(VECTOR (FORCE BALL WALL1 NORMAL :TIME 1) (DNUM 120 |deg|))

Quantities
IdNameSymbolEntries
(COMPO Y 0 (FORCE BALL WALL1 NORMAL :TIME 1))the y component of the normal force on the ball at T0 due to wall1Yc_Fn_BALL_WALL1_1_0
(GRAVITATIONAL-ACCELERATION EARTH)the gravitational acceleration due to the earthg_EARTH
(MASS BALL)the mass of the ballm_BALL
(MAG (FORCE BALL EARTH WEIGHT :TIME 1))the magnitude of the weight force on the ball at T0 due to the earthFw_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 earthYc_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 wall2Yc_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 wall2Fn_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 wall2Xc_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 earthXc_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 wall1Xc_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 wall1Fn_BALL_WALL1_1
Entries and Operators
EntryOperatorHints
(BODY BALL) (DRAW-BODY 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.
(DEFINE-VAR (GRAVITATIONAL-ACCELERATION EARTH)) (DEFINE-GRAV-ACCEL 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.
(DEFINE-VAR (MASS BALL)) (DEFINE-MASS BALL NIL)You can use the variable definition tools, which are under the variables menu, in order to define a variable for mass.
(DRAW-AXES 0) (DRAW-UNROTATED-AXES)Draw a standard horizontal-vertical coordinate system setting the positive x axis at 0 degrees.
(DRAW-VECTOR-ALIGNED-AXES 0)What would be a useful choice for the coordinate axes?Rotate the coordinate axes setting the x axis at 0 degrees.
(DRAW-PROJECTION-AXES 0)
(EQN (= |Fw_BALL_EARTH_1| (* |m_BALL| |g_EARTH|))) (WT-LAW 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|)))))) (COMPO-GENERAL-CASE 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 non-zero 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|)))))) (COMPO-GENERAL-CASE 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 non-zero 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)) (COMPO-PERPENDICULAR 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|)))))) (COMPO-GENERAL-CASE 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 non-zero 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|)))))) (COMPO-GENERAL-CASE 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 non-zero 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|))) (COMPO-PARALLEL-AXIS |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 non-zero 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|))) (WRITE-G-ON-EARTH)You should know the value of g for the EarthWrite the equation g_EARTH = 9.8 m/s^2
(EQN (= |m_BALL| (DNUM 2.0 |kg|))) (WRITE-KNOWN-VALUE-EQN (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)) (WRITE-NFL-COMPO 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)) (WRITE-NFL-COMPO 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
(IMPLICIT-EQN (= |OFn_BALL_WALL1_1| (DNUM 120 |deg|))) (WRITE-IMPLICIT-EQN (DIR (FORCE BALL WALL1 NORMAL :TIME 1)))
(IMPLICIT-EQN (= |OFn_BALL_WALL2_1| (DNUM 40 |deg|))) (WRITE-IMPLICIT-EQN (DIR (FORCE BALL WALL2 NORMAL :TIME 1)))
(IMPLICIT-EQN (= |OFw_BALL_EARTH_1| (DNUM 270 |deg|))) (WRITE-IMPLICIT-EQN (DIR (FORCE BALL EARTH WEIGHT :TIME 1)))
(IMPLICIT-EQN (= |Xc_Fw_BALL_EARTH_1_0| 0)) (COMPO-PARALLEL-AXIS |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 non-zero 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.
(IMPLICIT-EQN (= |Xc_a_BALL_1_0| 0)) (WRITE-NFL-COMPO 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
(IMPLICIT-EQN (= |Yc_a_BALL_1_0| 0)) (WRITE-NFL-COMPO 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
(IMPLICIT-EQN (= |a_BALL_1| (DNUM 0 |m/s^2|))) (WRITE-IMPLICIT-EQN (MAG (ACCEL BALL :TIME 1)))
(VECTOR (ACCEL BALL :TIME 1) ZERO) (ACCEL-AT-REST 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 zero-length acceleration vector for it.
(VECTOR (FORCE BALL EARTH WEIGHT :TIME 1) (DNUM 270 |deg|)) (DRAW-WEIGHT 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|)) (DRAW-NORMAL 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|)) (DRAW-NORMAL 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.