2Lesson 1 : Angular Position, Velocity, and Acceleration When a rigid object rotates about its axis, at any given time different parts of the object have different linear velocities and linear accelerations.A particle at point P at a fixed distance r from origin is rotating about axis O.
3P is at polar coordinate (r, q) q is measured counterclockwise from some reference line (q = 0)s = rqq =sr
4q is measured in radians (rad) One radian is the angle subtended by an arc length equal to the radius of the arc.q =sr360o =2prrrad = 2p rad1 rad =360o2p= 57.3o
5Converting from Degrees to Radians q (rad) =p180oq (deg)90o = p/2 rad60o = p/3 rad45o = p/4 rad270o = 3p/2 rad
6Angular Displacement (Dq) As particle moves from A to B in time interval Dt, the reference line of length r sweeps out an angleDq = qf - qi
7Average Angular Speed (w) Ratio of the angular displacement of a rigid object to the time interval Dt.w =qf - qitf - ti=DqDtThe rad/s is the unit for angular speed.w is positive when q increases (counterclockwise motion)w is negative when q decreases (clockwise motion)
9Instantaneous Angular Acceleration (a) a = limDt 0DwDt=dtdwThe rad/s2 is the unit for angular speed.a is positive when object rotates counterclockwise and speeds upORwhen object rotates clockwise and slows down
10Direction of Angular Velocity and Acceleration Vectors w and a are vector quantities with magnitude and directionThe directions of w and a are along the axis of rotation.
11Wrap four fingers of the right hand in the direction of rotation. Right-Hand RuleWrap four fingers of the right hand in the direction of rotation.Thumb will point in the direction of angular velocity vector (w).
12Example 1A rigid object is rotating with an angular speed w < 0. The angular velocity vector w and the angular acceleration vector a are antiparallel. The angular speed of the rigid object isa) clockwise and increasingb) clockwise and decreasingc) counterclockwise and increasingd) counterclockwise and decreasing
13Example 2During a certain period of time, the angular position of a swinging door is described by q = t t2, where q is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door ata) at t = 0b) at t = 3.00 s
14Lesson 2 : Rotational Kinematics with Constant Angular Acceleration dwdtdw = a dt(by integrating)wf = wi + atSince w =dqdtdqdtwi + at =
15qf = qi + wit + ½ at2 (by integrating) Eliminating t from previous two equations,Eliminating a from previous two equations,wf2 = wi2 + 2a(qf – qi)qf = qi + ½ (wi + wf) t
17A wheel rotates with a constant angular acceleration of 3.50 rad/s2. Example 1A wheel rotates with a constant angular acceleration of 3.50 rad/s2.a) If the angular speed of the wheel is rad/s at ti = 0, through what angular displacement does the wheel rotate in 2.00 s ?
18b) Through how many revolutions has. the wheel turned during this time b) Through how many revolutions has the wheel turned during this time interval ?c) What is the angular speed of the wheel at t = 2.00 s ?
19Example 2A rotating wheel requires 3.00 s to rotate through 37.0 revolutions. Its angular speed at the end of the 3.00 s interval is 98.0 rad/s. What is the constant angular acceleration of the wheel ?
20Lesson 3 : Angular and Linear Quantities Every particle of a rigid rotating object moves in a circle whose center is the axis of rotation.Tangential speed v =dsdtSince s = rq,v =dqdtrv = rw
21perpendicular distance of a point from the axis of rotation Tangential Speedv = rwangular speedperpendicular distance of a point from the axis of rotationTangential speed depends on distance from axis of rotationAngular speed is the same for all points
22Tangential Acceleration dvdtSince v = rw,at =dwdtrat = ra
23Centripetal Acceleration in terms of Angular Speed v2rSince v = rw,ac = rw2
24Total Linear Acceleration a = at2 + ar2a = r2a2 + r2w4a = r a2 + w4
25Example 1In order to keep the tangential speed of the disc surface at the location of the lens constant, the disc’s angular speed must vary as the lens moves radially along the disc. In a typical compact disc player, the constant speed of the surface at the point of the laser-lens system is 1.3 m/s.
26a) Find the angular speed of the disc in rev/min a) Find the angular speed of the disc in rev/min when information is being read from the innermost first track (r = 23 mm) and the outermost final track (r = 58 mm).
27b) The maximum playing time of a standard music. CD is 74 min 33 s b) The maximum playing time of a standard music CD is 74 min 33 s. How many revolutions does the disc make during that time ?
28c) What total length of track moves past the lens during this time ?
29d) What is the angular acceleration of the CD over d) What is the angular acceleration of the CD over the 4,473 s time interval ? [Assume that a is constant.]
30Example 2The drive train of a bicycle is shown to the right. The wheels are 67.3 cm in diameter and pedal cranks 17.5 cm long. The cyclist pedals at a steady angular rate of 76.0 rev/min. The chain engages with a front sprocket 15.2 cm in diameter and a rear sprocket 7.00 cm in diameter.
31a) Calculate the speed of a link of the a) Calculate the speed of a link of the chain relative to the bicycle frame.b) Calculate the angular speed of the bicycle wheels.
32c) Calculate the speed of the bicycle relative to the road. d) What pieces of data, if any, are not necessary for the calculations ?
33Lesson 4 : Rotational Kinetic Energy The total kinetic energy of a rotating rigid object is the sum of the kinetic energies of its individual particles.KErot = SKEi = S ½mivi2Since v = rw,KErot = ½ S miri2wi2Factoring out w2,KErot = ½ (S miri2) w2
34The kg . m2 is the SI unit for moment of inertia. KErot = ½ (S miri2) w2Moment of Inertia (I)I = S miri2The kg . m2 is the SI unit for moment of inertia.Moment of inertia is a measure of the resistance of an object to changes in its rotational motion.
35Substituting I,KErot = ½ Iw2Rotational MotionLinear MotionImwvKErot = ½ Iw2KE = ½ mv2
36Example 1Consider an oxygen molecule (O2) rotating in the x-y plane about the z-axis. The rotation axis passes through the center of the molecule, perpendicular to its length. The mass of each oxygen atom is 2.66 x kg, and at room temperature the average separation between the two atoms is d = 1.21 x m. (The atoms are modeled as particles.)a) Calculate the moment of inertia of the molecule about the z-axis.
37b) If the angular speed of the molecule about the. z-axis is 4 b) If the angular speed of the molecule about the z-axis is 4.60 x 1012 rad/s, what is its rotational kinetic energy ?
38Lesson 5 : Calculation of Moments of Inertia Moment of inertia of a rigid object is evaluated by dividing the object into many small volume elements, each with mass = Dmi.òI = lim Sri2Dmi =r2 dmDmi 0Since r = m/V,dm = r dVòI =rr2 dV
39Example 1Find the moment of inertia of a uniform thin hoop of mass M and radius R about an axis perpendicular to the plane of the hoop and passing through its center.
40Example 2Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an axis perpendicular to the rod (the y-axis) and passing through its center of mass.
41Example 3A uniform solid cylinder has a radius R, mass M, and length L. Calculate its moment of inertia about its central axis (the z-axis).
42Parallel-Axis Theorem The moment of inertia about any axis parallel to and a distance D away from this axis isI = ICM + MD2
43Example 4Consider once again the uniform rigid rod of mass M and length L. Find the moment of inertia of the rod about an axis perpendicular to the rod through one end (the y’ axis).
44Moment of Inertia of a Thin Cylindrical Shell (Hoop)
52The component Fsinq tends to rotate the wrench about point O. Lesson 6 : TorqueThe tendency of a force to rotate an object about some axis is measured by a vector quantity called torque.The component Fsinq tends to rotate the wrench about point O.
53Torque should not be confused with force. t = r Fsinf = Fddistance between pivot point and point of application of Fperpendicular distance from pivot point to the line of action of FTorque should not be confused with force.Torque has units of force x length or N.m. (Same as work but not called Joules.)
54F1 tends to rotate counterclockwise (+t) F2 tends to rotate clockwise (-t)St = t1 + t2 = F1d1 – F2d2The net torque about axis 0 is
55Example 1A one-piece cylinder is shaped as shown with a core section protruding from the larger drum. The cylinder is free to rotate about the central axis shown in the drawing. A rope wrapped around the drum, which has radius R1, exerts a force T1 to the right on the cylinder. A rope wrapped around the core, which has radius R2, exerts a force T2 downward on the cylinder.a) What is the net torque acting on the cylinder about the rotation axis (z-axis) ?
56b) Suppose T1 = 5. 0 N, R1 = 1. 0 m, T2 = 15. 0 N,. and R2 = 0. 50 m b) Suppose T1 = 5.0 N, R1 = 1.0 m, T2 = 15.0 N, and R2 = 0.50 m. What is the net torque about the rotation axis, and which way does the cylinder rotate starting from rest ?
57Lesson 7 : Relationship between Torque and Angular Acceleration Ft = matt = Ftr = (mat)rt = (mra)r = (mr2)aSince I = mr2,St = Ia(rotational analog to Newton’s Second Law)
58Example 1A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane. The rod is released from rest in the horizontal position. What is the initial angular acceleration of the rod and the initial linear acceleration of its right end ?
59Example 2A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless horizontal axis. A light cord wrapped around the wheel supports an object of mass m.a) Calculate the angular acceleration of the wheel.
60b) Calculate the linear acceleration of the object. c) Calculate the tension in the cord.
61Lesson 8 : Work, Power, and Energy in Rotational Motion Work done by F on an object as it rotates through an infinitesimal distance ds = r dqdW = F.ds = (Fsinf)r dq(The radial component of F does no work because it is perpendicular to the displacement.)
62(Power delivered to a rotating rigid object.) P = dW Since t = rFsinf,analogous to dW = Fx dx in linear motiondW = t dqRate at which work is being done as an object rotates through angle dq in time interval dt isdWdt=dqtanalogous to P = Fv in linear motion(Power delivered to a rotating rigid object.)P =dWdt= tw
64Integrating to find total work done, SW =òwiwfIw dw = ½ Iwf2 – ½ Iwi2Work–KE Theorem for Rotational MotionThe net work done by external forces in rotating a symmetric rigid object about a fixed axis equals the change in the object’s rotational energy.
66Example 1A uniform rod of length L and mass M is free to rotate on a frictionless pin passing through one end. The rod is released from rest in the horizontal position.a) What is its angular speed when it reaches its lowest position ?
67b) Determine the tangential speed of the b) Determine the tangential speed of the center of mass and the tangential speed of the lowest point on the rod when it is in the vertical position.
68Example 2Consider two cylinders having different masses m1 and m2, connected by a string passing over a pulley. The pulley has a radius R and moment of inertia I about its axis of rotation. The string does not slip on the pulley, and the system is released from rest. Find the linear speeds of the cylinders after cylinder 2 descends through a distance h, and the angular speed of the pulley at this time.
69Example 3 : AP 2001 #3A light string that is attached to a large block of mass 4m passes over a pulley with negligible rotational inertia and is wrapped around a vertical pole of radius r, as shown in Experiment A above. The system is released from rest, and as the block descends the string unwinds and the vertical pole with its attached apparatus rotates. The apparatus consists of a horizontal rod of length 2L, with a small block of mass m attached at each end. The rotational inertia of the pole and the rod are negligible.
70a) Determine the rotational inertia of the rod-and- a) Determine the rotational inertia of the rod-and- block apparatus attached to the top of the pole.b) Determine the downward acceleration of the large block.
71c) When the large block has descended a distance c) When the large block has descended a distance D, how does the instantaneous total kinetic energy of the three blocks compare with the value 4mgD ? Check the appropriate space below.____ Greater than 4mgD____ Equal to 4mgD____ Less than 4mgDJustify your answer.
72The system is now reset. The string is rewound around the pole to bring the large block back to its original location. The small blocks are detached from the rod and then suspended from each end of the rod, using strings of length l. The system is again released from rest so that as the large block descends and the apparatus rotates, the small blocks swing outward, as shown in Experiment B above. This time the downward acceleration of the block decreases with time after the system is released.
73d) When the large block has descended a distance d) When the large block has descended a distance D, how does the instantaneous total kinetic energy of the three blocks compare to that in part c) ? Check the appropriate space below.____ Greater____ Equal____ LessJustify your answer.
74Lesson 9 : Rolling Motion of a Rigid Object Cylinder rolling on a straight path.Center moves in a straight line (green line).A point on the rim moves in a path called a cycloid (red curve).
75Speed of CM of Cylinder Rolling without Slipping vCM =dsdtSince s = Rq,vCM =dqdtRvCM =Rwfor pure rolling motion only
76Acceleration of CM of Cylinder Rolling without Slipping aCM =dvCMdtaCM =dwdtRaCM =Ra
77Since IP = ICM + MR2, (parallel-axis theorem) Total Kinetic Energy of a Rolling CylinderKErot = ½ IPw2Since IP = ICM + MR2, (parallel-axis theorem)KErot = ½ (ICM + MR2)w2ORKErot = ½ ICMw2 + ½ MR2w2Since vCM = Rw,KErot = ½ ICMw2 + ½ MvCM2
78KEtotal = ½ ICMw2 + ½ MvCM2rotational KEtranslational KEThe total kinetic energy of a rolling object is the sum of the rotational kinetic energy about the center of mass and the translational kinetic energy of the center of mass.
80Example 1Three objects of uniform density – a solid sphere, a solid cylinder, and a hollow cylinder – are placed at the top of an incline. They are all released from rest at the same elevation and roll without slipping. Which object reaches the bottom first ? Which reaches it last ?
81Example 2For the solid sphere shown above, calculate the linear speed of the center of mass at the bottom of the incline and the magnitude of the linear acceleration of the center of mass.
82Example 3 : AP 1986 #2An inclined plane makes an angle of q with the horizontal, as shown above. A solid sphere of radius R and mass M is initially at rest in the position shown, such that the lowest point of the sphere is a vertical height h above the base of the plane. The sphere is released and rolls down the plane without slipping. The moment of inertia of the sphere about an axis through its center is 2/5 MR2. Express your answers in terms of M, R, h, g, and q.
83a) Determine the following for the sphere when it a) Determine the following for the sphere when it is at the bottom of the plane.i. Its translational kinetic energyii. Its rotational kinetic energy
84b) Determine the following for the sphere when it is on the plane. i. Its linear accelerationii. The magnitude of the frictional force acting on it
85The solid sphere is replaced by a hollow sphere of identical radius R and mass M. The hollow sphere, which is released from the same location as the solid sphere, rolls down the incline without slipping.c) What is the total kinetic energy of the hollow sphere at the bottom of the plane
86d) State whether the rotational kinetic energy of d) State whether the rotational kinetic energy of the hollow sphere is greater than, less than, or equal to that of the solid sphere at the bottom of the plane. Justify your answer.
87Lesson 10 : Angular Momentum SF =From linear motion :dpdtTake cross product of each side with r :r x SF = r xdpdtNet TorqueSt = r xdpdt
88Add x p to right-hand side : Analogous to dp in translational motion drdrx p is zero since = v,dtdtand v and p are parallel.Add x p to right-hand side :drdtSt = r xdp+x pAnalogous to dp in translational motionangular momentum (L)L = r x pSt =d(r x p)dt
89Instantaneous Angular Momentum L = r x pThe instantaneous angular momentum L of a particle relative to the origin O is defined by the cross product of the particle’s instantaneous position vector r and its instantaneous linear momentum p.The SI unit of angular momentum is kg . m2/s.
90rotational analog of Newton’s second law Since L = r x p,St =dLdtrotational analog of Newton’s second lawThe torque acting on a particle is equal to the time rate of change of the particle’s angular momentum.
91The direction of L is always perpendicular to the plane formed by r and p. Since p = mv,L = mvr sinf
92Example 1A particle moves in the xy plane in a circular path of radius r, as shown above. Find the magnitude and direction of its angular momentum relative to O when its linear velocity is v.
93Angular Momentum of a System of Particles The total angular momentum of a system of particles about some point is defined as the vector sum of the angular momenta of the individual particles.Ltot = L1 + L2 + …. + Ln = SLiDifferentiating with respect to time :dLtotdt= SdLiSt
94St =dLtotdtThe net external torque acting on a system about some axis passing through an origin in an inertial frame equals the time rate of change of the total angular momentum of the system about that origin.** This theorem applies even if the center of mass is accelerating, as long as t and L are evaluated relative to the center of mass.
95Example 2A sphere of mass m1 and a block of mass m2 are connected by a light cord that passes over a pulley, as shown above. The radius of the pulley is R, and the mass of the rim is M. The spokes of the pulley have negligible mass. The block slides on a frictionless, horizontal surface. Find an expression for the linear acceleration of the two objects, using the concepts of angular momentum and torque.
96Lesson 11 : Angular Momentum of a Rotating Rigid Object Each particle of this rigid object rotates in the xy plane about the z axis with angular speed w.Since v = rw,L = mr2wSince I = mr2,L = Iw
97Differentiating with respect to time : L = IwDifferentiating with respect to time :(I is constant for a rigid object)dLdt= Idw= IaSt =dLdtSincerotational form of Newton’s second lawSt = Ia
98Example 1Estimate the magnitude of the angular momentum of a bowling ball spinning at 10 rev/s.
99Example 2A father of mass mf and daughter of mass md sit on opposite ends of a seesaw at equal distances from the pivot at the center. The seesaw is modeled as a rigid rod of mass M and length l and is pivoted without friction. At a given moment, the combination rotates in a vertical plane with an angular speed w.
100a) Find an expression for the magnitude of the a) Find an expression for the magnitude of the system’s angular momentum.
101b) Find an expression for the magnitude of the b) Find an expression for the magnitude of the angular acceleration of the system when the seesaw makes an angle q with the horizontal.
102Example 3 : AP 1983 #2A uniform solid cylinder of mass m1 and radius R is mounted on frictionless bearings about a fixed axis through O. The moment of inertia of the cylinder about the axis is I = ½ m1R2. A block of mass m2, suspended by a cord wrapped around the cylinder as shown above, is released at time t = 0.
103a) On the diagram below draw and identify all of a) On the diagram below draw and identify all of the forces acting on the cylinder and on the block.
104b) In terms of m1, m2, R, and g, determine each of the following. i) The acceleration of the block.
105ii) The tension in the cord. iii) The angular momentum of the disk as a function of t.
106Example 4 : AP 1982 #3A system consists of two small disks, of masses m and 2m, attached to a rod of negligible mass of length 3l as shown above. The rod is free to turn about a vertical axis through point P. The two disks rest on a rough horizontal surface; the coefficient of friction between the disks and the surface is m. At time t = 0, the rod has an initial counterclockwise angular velocity wo about P. The system is gradually brought to rest by friction. Develop expressions for the following quantities in terms of m, m, l, g, and wo.
107a) The initial angular momentum of the system about the axis through P. b) The frictional torque acting on the system about the axis through P.
108c) The time T at which the system will come to rest.
109Consider a thin uniform rod of mass M and length l , as shown above. Example 5 : AP 1996 #3Consider a thin uniform rod of mass M and length l , as shown above.lMa) Show that the rotational inertia of the rod about an axis through its center and perpendicular to its length is Ml 2/12.
110The rod is now glued to a thin hoop of mass M and radius R = l /2 to form a rigid assembly, as shown above. The centers of the rod and hoop coincide at point P. The assembly is mounted on a horizontal axle through point P and perpendicular to the page.
111b) What is the rotational inertia of the rod-hoop b) What is the rotational inertia of the rod-hoop assembly about the axle ?
112Several turns of string are wrapped tightly around the circumference of the hoop. The system is at rest when a cat, also of mass M, grabs the free end of the string and hangs vertically from it without swinging as it unwinds, causing the rod-hoop assembly to rotate. Neglect friction and the mass of the string.c) Determine the tension T in the string.
113d) Determine the angular acceleration a of the rod-hoop assembly. e) Determine the linear acceleration of the cat.
114f) After descending a distance H = 5 l /3, the cat lets f) After descending a distance H = 5 l /3, the cat lets go of the string. At that instant, what is the angular momentum of the cat about point P ?
115Lesson 12 : Conservation of Angular Momentum The total angular momentum of a system is constant in both magnitude and direction if the resultant external torque acting on the system is zero, that is, if the system is isolated.St =dLtotdt= 0,SinceLtot = constantLbefore = Lafter
116Lbefore = Lafter Since L = Iw, Iiwi = Ifwf = constant When arms are moved inward, I decreases.Since Iw remains constant, as I decreases, w must increase.
117Example 1A horizontal platform in the shape of a circular disk rotates freely in a horizontal plane about a frictionless vertical axle. The platform has a mass M = 100 kg and a radius R = 2.0 m. A student whose mass is m = 60 kg walks slowly from the rim of the disk toward its center. If the angular speed of the system is 2.0 rad/s when the student is at the rim, what is the angular speed when he reaches a point r = 0.50 m from the center ?
118Example 2In a favorite classroom demonstration, a student holds the axle of a spinning bicycle wheel while seated on a stool that is free to rotate. The student and stool are initially at rest while the wheel is spinning in a horizontal plane with an initial angular momentum Li that points upward. When the wheel is inverted about its center by 180o, the student and stool start rotating. In terms of Li, what are the magnitude and the direction of L for the student plus stool ?
119Example 3A 2.0 kg disk traveling at 3.0 m/s strikes a 1.0 kg stick of length 4.0 m that is lying flat on nearly frictionless ice. Assume that the collision is elastic and that the disk does not deviate from its original line of motion. Find the translational speed of the disk, the translational speed of the stick, and the angular speed of the stick after the collision. The moment of inertia of the stick about its center of mass is 1.33 kg.m2.
120Example 4 : AP 1987 #3A 1.0 kg object is moving horizontally with a velocity of 10 m/s, as shown above, when it makes a glancing collision with the lower end of a bar that was hanging vertically at rest before the collision. For the system consisting of the object and bar, linear momentum is not conserved in this collision, but kinetic energy is conserved. The bar, which has a length l = 1.2 m and a mass m = 3.0 kg, is pivoted about the upper end.
121Immediately after the collision the object moves with speed v at an angle q relative to its original direction. The bar swings freely, and after the collision reaches a maximum angle of 90o with respect to the vertical. The moment of inertia of the bar about the pivot is Ibar = ml 2/3. Ignore all friction.a) Determine the angular velocity of the bar immediately after the collision.
122b) Determine the speed v of the 1 kg object b) Determine the speed v of the 1 kg object immediately after the collision.c) Determine the magnitude of the angular momentum of the object about the pivot just before the collision.
124Example 5 : AP 1992 #2Two identical spheres, each of mass M and negligible radius, are fastened to opposite ends of a rod of negligible mass and length 2l. This system is initially at rest with the rod horizontal, as shown above, and is free to rotate about a frictionless, horizontal axis through the center of the rod and perpendicular to the plane of the page. A bug, of mass 3M, lands gently on the sphere on the left. Assume that the size of the bug is small compared to the length of the rod. Express your answers to all parts of the question in terms of M, l , and physical constants.
125a) Determine the torque about the axis immediately a) Determine the torque about the axis immediately after the bug lands on the sphere.b) Determine the angular acceleration of the rod- spheres-bug system immediately after the bug lands.
126The rod-spheres-bug system swings about the axis The rod-spheres-bug system swings about the axis. At the instant that the rod is vertical, as shown above, determine each of the following.c) The angular speed of the bug.
127d) The angular momentum of the system. e) The magnitude and direction of the force that must be exerted on the bug by the sphere to keep the bug from being thrown off the sphere.
128Lesson 13 : Rotational Equilibrium A system is in rotational equilibrium if the net torque on it is zero about any axis.St = 0Since St = Ia,a = 0St = 0 does not mean an absence of rotational motion. Object can be rotating at a constant angular speed.
129Example 1Consider the object subject to the two forces shown to the right. Choose the correct statement with regard to this situation. The object is in____ force equilibrium but not torque equilibrium.____ torque equilibrium but not force equilibrium.____ both force and torque equilibrium.____ neither force nor torque equilibrium.
130Example 2Consider the object subject to the three forces shown to the right. Choose the correct statement with regard to this situation. The object is in____ force equilibrium but not torque equilibrium.____ torque equilibrium but not force equilibrium.____ both force and torque equilibrium.____ neither force nor torque equilibrium.
131Center of gravity = center of mass if g is constant over the object. To compute the torque due to the gravitational force on an object of mass M, we need only consider the force Mg acting at the center of gravity of the object.Center of gravity = center of mass if g is constant over the object.
132Example 3A seesaw consisting of a uniform board of mass M and length l supports a father and daughter with masses mf and md, respectively. The support (called the fulcrum) is under the center of gravity of the board, the father is a distance d from the center, and the daughter is a distance l /2 from the center.
133a) Determine the magnitude of the upward force n a) Determine the magnitude of the upward force n exerted by the support on the board.b) Determine where the father should sit to balance the system.
134Example 4A person holds a 50.0 N sphere in his hand. The forearm is horizontal, as shown above. The biceps muscle is attached 3.00 cm from the joint, and the sphere is 35.0 cm from the joint. Find the upward force exerted by the biceps on the forearm and the downward force exerted by the upper arm on the forearm and acting at the joint. Neglect the weight of the forearm.