PHY 111 Complete Lecture Note PDF

Mechanics and Properties of Matter Module V & VI Presenter: Dr R. Sule OUTLINE Conservative forces, Conservation of linear momentum, Kinetic energy and Work, Potential energy, System of particles, Centre of mass, Torque, Vector product, Moment ...

Mechanics and Properties of Matter Module V & VI Presenter: Dr R. Sule OUTLINE Conservative forces, Conservation of linear momentum, Kinetic energy and Work, Potential energy, System of particles, Centre of mass, Torque, Vector product, Moment  DEFINITION OF A CONSERVATIVE FORCE Version 1 A force is conservative when the work it does on a moving object is independent of the path between the object’s initial and final positions. Wgravity = mg (ho − h f ) Version 2 A force is conservative when it does no work on an object moving around a closed path, starting and finishing at the same point. Wgravity = mg (ho − h f ) ho = h f The figure illustrate a roller coaster car racing through dips and double dips, ultimately returning to its starting point. A roller coaster track is an example of a closed path. ❖Gravity provides the only force that does work on the car, assuming that there is no friction or air resistance. ❖On the downward parts of the trip, the gravitational force does positive work increasing the car’s kinetic energy. ❖On the upward parts of the motion, the gravitational force does negative work, decreasing the car’s kinetic energy. ❖The gravitational force does as much positive work as negative work, so the net work is zero: 𝑊gravity = 0 𝐽 ❖Example of a conservative force: (i) Gravitational force (ii) Elastic force of a spring and (iii) electrical force of electrically charged particles  LINEAR MOMENTUM AND COLLISION There are many situations when the force on an object is not constant. 1(a) (Picture: Cutnel and Johnson 9th Edition) Figure 1a shows a baseball being hit, and part b of the figure illustrates approximately how the force applied to the ball by the bat changes during the time of contact. tf DEFINITION OF IMPULSE The impulse of a force is the product of the average force and the time interval during which the force acts:   J = F t Impulse is a vector quantity and has the same direction as the average force. newton  seconds (N  s)  DEFINITION OF LINEAR MOMENTUM The linear momentum 𝑷 of an object is the product of the object’s mass 𝒎 times its velocity 𝑽 :   p = mv Linear momentum is a vector quantity that points in the same direction as the velocity. SI Unit of Linear Momentum: kilogram. meter/second (kg. m/s)     vf − vo a= Eq. 1 t    F = ma  mv f − mv o  F = t Eq. 2 ( )     F t = mvf − mvo Eq. 3 The equation 3 is called impulse–momentum theorem  IMPULSE-MOMENTUM THEOREM When a net force acts on an object, the impulse of this force is equal to the change in the momentum of the object Ԧ During a collision, it is often difficult to measure the net average force σ 𝐹, so it is not easy to determine the impulse (σ 𝐹)∆𝑡Ԧ directly impulse ( )    F t = mvf − mvo final momentum initial momentum  Application During the launch of the space shuttle, the engines fire and apply an impulse to the shuttle and the launch vehicle. This impulse causes the momentum of the shuttle and the launch vehicle to increase, and the shuttle rises to its orbit. Example 1 A baseball (m = 0.14 kg) has an initial velocity of 𝑉𝑜 = −38 m/s as it approaches a bat. We have chosen the direction of approach as the negative direction. The bat applies an average force 𝐹തԦ that is much larger than the weight of the ball, and the ball departs from the bat with a final velocity of 𝑉𝑓 = + 58 m/s. (a) Determine the impulse applied to the ball by the bat. (b) Assuming that the time of contact is ∆t = 1.6 × 10−3 s, find the average force exerted on the ball by the bat. Solution: ❖ Two forces act on the ball during impact, and together they constitute the net average force: the average force exerted by the bat 𝐹, and the weight of the ball. ❖ Since 𝐹തԦ is much greater than the weight of the ball, we neglect the weight. Thus, the net average force is equal to 𝐹തԦ or σ 𝐹തԦ = 𝐹തԦ Recall, ( )    F t = mvf − mvo = +𝟏𝟑. 𝟒𝟒 𝒌𝒈. 𝒎/𝒔   J = F t Recall, = +𝟖𝟒𝟎𝟎𝑵 Example 2 During a storm, rain comes down with a velocity of -15 m/s and hits the roof of a car perpendicularly. The mass of rain per second that strikes the roof of the car is 0.060 kg/s. Assuming that rain comes to rest upon striking the car, find the average force exerted by the rain on the roof. ❖ Neglecting the weight of the raindrops, the net force on a raindrop is simply the force on the raindrop due to the roof.      m  F t = mv f − mv o F = −  v o  t   F = −(0.060 kg s )(− 15 m s ) = +0.90 N Class activity 1 Jennifer is walking at 1.63 m/s. If Jennifer weighs 583 N, what is the magnitude of her momentum? Class activity 2 A 1.0-kg ball has a velocity of 12 m/s downward just before it strikes the ground and bounces up with a velocity of 12 m/s upward. What is the change in momentum of the ball? Solution: class activity 1 Recall P = mv V = 1.63 m/s W = m×g 583 N = m × 9.8 m/s2 m = 59.49 g P = 59.49 × 1.63 = 96.97 kg. m/s Solution : class activity 2 Change in momentum = final momentum – initial momentum 𝑚𝐯𝐟 − 𝑚𝐯𝐨 𝐯𝐟 : upward and 𝐯𝐨 : downward (+ 1 × 12 upward) – (- 1 ×12 downward) 24 kg. m/s upward  Class activity 3 and solution A projectile is launched with 200 kg  m/s of momentum and 1000 J of kinetic energy. What is the mass of the projectile? Solution: 1 Kinetic energy = 𝑚𝑣 2 , P = mv then, substitute p for mv 2 1 K.E = 𝑃𝑣= 1000 J 2 V = 2000/200 = 10 P = mv M = 200/10 = 20 kg Mass = 20kg Class activity 4 A 1055-kg van, stopped at a traffic light, is hit directly in the rear by a 715-kg car traveling with a velocity of +2.25 m/s. Assume that the transmission of the van is in neutral, the brakes are not being applied, and the collision is elastic. What is the final velocity of the car. Solution: class activity 1 The final velocity 𝒗𝒇𝟏 of the car is given by the equation 𝒎𝟏 −𝒎𝟐 𝒗𝒇𝟏 = 𝒗𝟎𝟏 𝒎𝟏 +𝒎𝟐 where 𝑚1 and 𝑚2 are respectively, the masses of the car and the van, and 𝑣01 is the initial of the car. kg−𝟏𝟎𝟓𝟓 kg 𝟕𝟏𝟓 Thus, 𝒗𝒇𝟏 = +𝟐. 𝟐𝟓 m/s = −𝟎. 𝟒𝟑𝟐 𝒎/𝒔 𝟕𝟏𝟓 kg+𝟏𝟎𝟓𝟓 kg What is the final velocity of the car = -0.432 m/s Assignment: Question 1 Two objects have the same momentum. Do the velocities of these objects necessarily have (a) the same directions and (b) the same magnitudes? Give your reasoning in each case. Question 2 A volleyball is spiked so that its incoming velocity of + 4.0 m/s is changed to an outgoing velocity of – 21 m/s. The mass of the volleyball is 0.35 kg. What impulse does the player apply to the ball?  The Principle of Conservation of Linear Momentum ❖We apply the impulse-momentum theorem to the midair collision between two objects. ❖The two objects (masses m1 and m2) are approaching each other with initial velocities V01 and V02, as shown in the Fig. (a). ❖They interact during the collision in part b and then depart with the final velocities Vf1 and Vf2 as shown in part (c). ❖Because of the collision, the initial and final velocities are not the same. Two types of forces act on the system Internal forces – Forces that objects within the system exert on each other. External forces – Forces exerted on objects by agents external to the system. During the collision in part (b) 𝐹12 is the force exerted on object 1 by object 2, while 𝐹21 is the force exerted on object 2 by object 1. These forces are action–reaction forces, 𝐹12 = − 𝐹21  Principle of Conservation of Linear Momentum ( )     F t = mv f − mv o OBJECT 1 (  )   W1 + F12 t = m1vf 1 − m1vo1 OBJECT 2 (  )   W2 + F21 t = m2 vf 2 − m2 vo 2 The force of gravity also acts on the objects, their weights 𝑊1 𝑎𝑛𝑑 𝑊2 : External forces  (   )   W1 + F12 t = m1vf 1 − m1vo1 + (   )   W2 + F21 t = m2 vf 2 − m2 vo 2 (     )     W1 + W2 + F12 + F21 t = (m1v f 1 + m2 vf 2 ) − (m1vo1 + m2 vo 2 )     F12 = − F21 Pf Po  The internal forces cancel out because 𝐹12 = − 𝐹21 ( )     W1 + W2 t = Pf − Po   (sum of average external forces)t = Pf − Po ❖ For an isolated system, the sum of the external forces is zero. i.e W1 + W2 = 0 ❖ If the sum of the external forces is zero, then     0 = Pf − Po Pf = Po The above equation show that total linear momentum of an isolated system is constant or conserved. Example A freight train is being assembled in a switching yard as shown in the diagram. Car 1 has a mass of m1 = 65 × 103 kg and moves at a velocity of v01 = + 0.80 m/s. Car 2, with a mass of m2 = 92 × 103 kg and a velocity of v02= 1.3 m/s, overtakes car 1 and couples to it. Neglecting friction, find the common velocity vf of the cars after they become coupled. = 𝑚1 + 𝑚2 𝐯𝐟 = 𝑚1 𝐯𝐨1 + 𝑚2 𝐯𝐨2 Total momentum after collision Total momentum before collision (65× 103kg)(0.80 m/s)+(92 ×103kg)(1.3 m/s) 𝑣𝑓 = = 1.09 m/s (65×103 kg + 92×103 kg) Class activity Starting from rest, two skaters push off against each other on ice where friction is negligible. One is a 54-kg woman and one is a 88-kg man. The woman moves away with a speed of +2.5 m/s. Find the recoil velocity of the man.    Pf = Po m1v f 1 + m2 v f 2 = 0 m1v f 1 vf 2 = − m2 54 kg +2.5 mΤs 𝑣𝑓2 =− = −1.53 mΤs 88 kg The total momentum of the skaters before they push on each other is zero, since they are at rest.  The total linear momentum is conserved when two objects collide, provided they constitute an isolated system. Elastic collision -- One in which the total kinetic energy of the system after the collision is equal to the total kinetic energy before the collision. Inelastic collision -- One in which the total kinetic energy of the system after the collision is not equal to the total kinetic energy before the collision; if the objects stick together after colliding, the collision is said to be completely inelastic. Example an elastic head-on collision between two balls. One ball has a mass of m1 = 0.250 kg and an initial velocity of 5.00 m/s. The other has a mass of m2 = 0.800 kg and is initially at rest. No external forces act on the balls. What are the velocities of the balls after the collision? 1 2 1 2 1 2 𝑚1 𝑣𝑓1 + 𝑚2 𝑣𝑓2 = 𝑚1 𝑣01 +0 2 2 2 2 Make 𝑣𝑓1 the subject of formula 𝑅𝑒𝑐𝑎𝑙𝑙 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑚1 𝐯𝐟1 + 𝑚2 𝐯𝐟2 = 𝑚1 𝐯𝐨1 + 𝑚2 𝐯𝐨2 Make 𝐯𝐟2 the subject of formula and substitute it into kinetic energy equation. Solution: Since the collision is elastic, the kinetic energy of the two-ball system is the same before and after the balls collide: 1 2 1 2 1 2 𝑚1 𝑣𝑓1 + 𝑚2 𝑣𝑓2 = 𝑚1 𝑣01 +0 2 2 2 The total kinetic energy after collision = The total kinetic energy before collision 𝑚1 − 𝑚2 𝑣𝑓1 = 𝑣01 𝑚1 + 𝑚2 0.25𝑘𝑔 − 0.800𝑘𝑔 × 5 𝑚/𝑠 = −2.62𝑚/𝑠 0.250𝑘𝑔 + 0.800 𝑘𝑔 2𝑚1 2 (0.25𝑘𝑔) 𝑣𝑓2 = 𝑣01 = × 5 m/s = +2.38 m/s 𝑚1 +𝑚2 0.250𝑘𝑔+0.800 𝑘𝑔 The negative value for vf1 indicates that ball 1 rebounds to the left after the collision Applying the Principle of Conservation of Linear Momentum 1. Decide which objects are included in the system. 2. Relative to the system, identify the internal and external forces. 3. Verify that the system is isolated. 4. Set the final momentum of the system equal to its initial momentum. Remember that momentum is a vector.  Work and Energy ❖ Work provides a link between force and energy ❖ The work ‘W’ done by a constant force on an object is defined as the product of components of force along the direction of displacement and the magnitude of displacement. ❑ F is the magnitude of the force ❑ ∆𝑋 is the magnitude of the object’s displacement ❑ 𝜃 is the angle between 𝐅Ԧ and Δ𝐱  W = Fs 1 N  m = 1 joule (J ) Work is a scalar quantity ❖ The work done by a force is zero Displacement is when the force is perpendicular to the horizontal displacement Force is vertical cos 90° = 0 cos 90° = 0 ❖ If there are multiple forces acting on an object, the total work done is the algebraic sum of the amount of work done by each force Work is Zero  cos 0 = 1 W = (F cos )s cos90 = 0 cos180 = −1 Example 1 Pulling a Suitcase-on-Wheels Find the work done if the force is 45.0-N, the angle is 50.0 degrees, and the displacement is 75.0 m.   W = (F cos )s = (45.0 N ) cos 50.0 (75.0 m ) = 2170 J ❖ Work can be positive or negative ❖ Positive if the force and displacement are in the same direction (b) ❖ Negative if the force and displacement are in the opposite direction (c) W = (F cos 0)s = Fs W = (F cos180)s = −Fs Example: Accelerating a Crate The truck is accelerating at a rate of +1.50 m/s2. The mass of the crate is 120-kg and it does not slip. The magnitude of the displacement is 65 m. What is the total work done on the crate by all of the forces acting on it? The angle between the displacement and the normal force is 90 degrees. The angle between the displacement and the weight is also 90 degrees. W = (F cos90)s = 0 The angle between the displacement and the friction force is 0 degrees. ( ) f s = ma = (120 kg ) 1.5 m s 2 = 180N W = (180N ) cos 0(65 m ) = 1.2 104 J  Work-Energy Theorem and Kinetic Energy Consider a constant net external force acting on an object. The object is displaced a distance s, in the same direction as the net force. F s The work is simply W = ( F )s = (ma )s  𝑊 = 𝐹𝑠 = 𝑚 𝑎𝑠 𝑣𝑓2 = 𝑣𝑜2 + 2 𝑎𝑠 1 2 𝑎𝑠 = 𝑣𝑓 − 𝑣𝑜2 2 1 2 𝑊 = 𝐹𝑠 = 𝑚 𝑎𝑠 = 𝑚 𝑣𝑓 − 𝑣𝑜2 2 1 2 1 𝑊 = 𝑚𝑣𝑓 − 𝑚𝑣𝑜2 2 2 The above expression is the work-energy theorem i.e the left is work done and the right is kinetic energy. DEFINITION OF KINETIC ENERGY The kinetic energy KE of an object with mass m and speed v is given by KE = 1 𝑚𝑣 2 2 ❖Energy associated with the motion of an object kinetic energy KE ❖Is a scalar quantity with the same units as work (J). When work is done by a net force on an object and the only change in the object is its speed, the work done is equal to the change in the object’s kinetic energy Wnet = KEf − KEi = KE  Example: work-energy theorem When a net external force does work on and object, the kinetic energy of the object changes according to W = KEf − KEo = mv − mv 1 2 2 f 1 2 2 o Deduction from the figure above: The work-energy theorem does not apply to the work done on an individual force, unless that force happen to be the only one present, in that case is a net force. If the work done by the net force is positive, as in example above, the kinetic energy of the object increases. If the work done is negative, the kinetic energy decreases. If the work is zero, the kinetic energy remains the same. Example 1 The mass of the space probe is 474-kg and its initial velocity is 275 m/s. If the 56.0-mN force acts on the probe through a displacement of 2.42×109m, what is its final speed?  ( F)cos s = 1 2 mvf2 − 12 mvo2 (5.6010 N)cos0 (2.4210 m) = (474 kg )v -2  9 1 2 2 f − 12 (474 kg )(275m s ) 2 v f = 805m s GRAVITATIONAL POTENTIAL ENERGY The gravitational potential energy PE is the energy that an object of mass m has by virtue of its position relative to the surface of the earth. That position is measured by the height h of the object relative to an arbitrary zero level: PE = mgh 1 N  m = 1 joule (J )  W = (F cos )s Wgravity = mg (ho − h f ) Example: The gymnast leaves the trampoline at an initial height of 1.20 m and reaches a maximum height of 4.80 m before falling back down. What was the initial speed of the gymnast?  W = 12 mvf2 − 12 mvo2 mg (ho − h f ) = − 12 mvo2 Wgravity = mg (ho − h f ) vo = − 2 g (ho − h f ) ( ) vo = − 2 9.80 m s 2 (1.20 m − 4.80 m) = 8.40 m s  CENTER OF MASS The center of mass is a point that represents the average location for the total mass of a system. m1 x1 + m2 x2 xcm = m1 + m2 1 If m1 = m2 = m, the above Equation becomes 𝑥𝑐𝑚 = 𝑥1 + 𝑥2 which 2 corresponds to the point midway between the particles.  Example Suppose that 𝑚1 = 5.0 kg and 𝑥1 = 2.0 m, while 𝑚2 = 12 kg and 𝑥2 = 6.0 m. Then we expect the average location of the total mass to be located closer to particle 2, since it is more massive. m1 x1 + m2 x2 xcm = m1 + m2 5 × 2 + (12 × 6) 𝑥𝑐𝑚 = = 4.82 𝑚 5 + 12 Hence, the average location of the total mass is located closer to particle 2, since it is more massive. If a system contains more than two particles, the center-of-mass point can be determined by generalizing Equation.  SYSTEM OF PARTICLES When dealing with large number of particles, it is convenient and essential to describe the motion in the center of mass coordinates. Consider a system containing 𝑁 particles labeled 1, 2, 3, ……, 𝑁 with masses 𝑚1 , 𝑚2 , 𝑚3 , …. , 𝑚𝑁 located at position vector 𝑟1 , 𝑟2 , 𝑟3 , … … … , 𝑟𝑁 from the origin O as shown in the figure below. The center of mass or centroid of the system of particles is defined as that point C having position vector ; 𝑁 𝑚1 𝑟1 + 𝑚2 𝑟2 + ⋯ … ….. 𝑚𝑛 𝑟𝑛 1 σ 𝑚𝑣 𝑟𝑣 𝑟ሶ = = ෍ 𝑚𝑣 𝑟𝑣 = …………..1 𝑚1 + 𝑚2 + ⋯ … … 𝑚𝑛 𝑀 𝑀 𝑣=1 𝑁 𝑀 = ෍ 𝑚𝑣 is the sum of all the mass in the system. 𝑣=1 In component form we can write 1 1 1 𝑋 = ෍ 𝑚𝑣 𝑥𝑣 , 𝑌 = ෍ 𝑚𝑣 𝑦𝑣 , 𝑍 = ෍ 𝑚𝑣 𝑧𝑣 𝑀 𝑀 𝑀 By differentiating equation 1, we obtained the velocity of the center of mass as: σ 𝑚𝑣 𝑥ሶ 𝑣 𝑉𝑥 = 𝑋ሶ = 𝑀 The acceleration of the center of the mass is obtain by differentiating the velocity. Example A system of particles consist of a 3 gram mass located at (1, 0, -1), a 5 gram mass at (-2, 1, 3) and a 2 gram mass at (3, -1, 1). Find the coordinates of the center of mass. Soln: The position vectors of the particles are given respectively; 𝑟1 = 𝑖 − 𝑘, 𝑟2 = −2𝑖 + 𝑗 + 3𝑘, 𝑟3 = 3𝑖 − 𝑗 + 𝑘 𝑚1 𝑟1 +𝑚2 𝑟2 +𝑚3 𝑟3 Then, 𝑟ሶ = 𝑚1 +𝑚2 +𝑚3 3 𝑖−𝐾 +5 −2𝑖+𝑗+3𝑘 +2(3𝑖−𝑗+𝑘) −1 3 14 𝑟ሶ = = 𝑖+ 𝑗+ 𝑘 3+5+2 10 10 10 −1 3 14 Thus, the coordinate of the center of mass are ( , , ) 10 10 10  DEFINITION OF TORQUE ❖ When a force is exerted on a rigid object pivoted about an axis, the object tends to rotate about that axis. ❖ The tendency of a force to rotate an object about some axis is measured by a vector quantity called torque. Magnitude of Torque = (Magnitude of the force) x (Lever arm) = 𝜏 = 𝐹ℓ The lever arm is the distance ℓ between the line of action and the axis of rotation, measured on a line that is perpendicular to both. ❖ Direction: The torque is positive when the force tends to produce a counterclockwise rotation about the axis. ❖ SI Unit of Torque: newton x meter (N·m) 𝜏 = 𝐹ℓ indicates that forces of the same magnitude can produce different torques, depending on the value of the lever arm. Example 1 A force (magnitude = 55 N) is applied to a door. However, the lever arms are different in the three doors (a) 0.80 m, (b) 0.60 m, and (c) 0 m. Find the torque in each case. Answer (a) 𝜏 = 𝐹ℓ = (55 N)(0.80 m) = +44 N.m (b) 𝜏 = 𝐹ℓ = (55 N)(0.60 m) = +33 N.m (c) 𝜏 = 𝐹ℓ = (55 N)(0 m) = ) N.m Example 2 The Figure shows the ankle joint and the Achilles tendon attached to the heel at point P. The tendon exerts a force of magnitude 790 N. Determine the torque (magnitude and direction) of this force about the ankle joint which is located 3.6 × 10−2 𝑚 away from point P. Solution To calculate the magnitude of the torque, it is necessary to have a value for the lever arm. The lever arm is the perpendicular distance between the axis of rotation at the ankle joint and the line of action of the force.  The lever arm is ℓ  cos 55 =  3.6 10− 2 m  = F 790 N The magnitude of the torque is  = (720 N )(3.6  10−2 m )cos 55 = 15 N  m The force 𝐹 tends to produce a clockwise rotation about the ankle joint, so the torque is negative: 𝜏 = −15 N. m  CROSS PRODUCT AND TORQUE If a force F acted on a rigid object as shown in Figure below. The effect of the force depends on the location of its point of application P. If r is the position vector of this point relative to O, the torque associated with the force F about O is given by: 𝜏 =𝑟×𝐹 ❖ The torque is related to the position vector r and force F mathematically by a vector product: 𝜏 =𝑟×𝐹 We shall not treat cross product in this module because it has been thought under vector. Further reading (Physics for Scientist and Engineers. Activity A force of 𝐹 = 2𝑖 + 3𝑗 𝑁 is applied to an object that is pivoted about a fixed axis aligned along the z-component axis. If the force is applied at a point located at r = 4𝑖 + 5𝑗 𝑚, find the torque.  MOMENT ❖ A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. In equilibrium, the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero.  Fx = 0  Fy = 0  =0  Balanced Rock in Arches National Park, Utah (Physics for scientist and Engineering) Example 3 A woman whose weight is 530 N is poised at the right end of a diving board with length 3.90 m. The board has negligible weight and is supported by a fulcrum 1.40 m away from the left end. Find the forces that the bolt and the fulcrum exert on the board. Solution Three forces act on the board: F1 , F2 and W 𝐅𝟏 points downward because the bolt must pull in that direction to counteract the tendency of the board to rotate clockwise about the fulcrum. 𝐅𝟐 points upward, because the board pushes downward against the fulcrum, which, in reaction, pushes upward on the board. ❖ Since the board is in equilibrium, the sum of the vertical forces must be zero: σ 𝐹𝑦 = 𝐹1 + 𝐹2 + 𝑊 = 0 ❖ Also, the sum of the torques must be zero: σ 𝜏 = 0 ❖ To calculating torques, we select an axis that passes through the left end of the board The force 𝐹1 produces no torque because it has a zero lever arm, while 𝐹2 creates a counterclockwise (positive) torque, and 𝑊 produces a clockwise (negative) torque  = F  2 2 − W W = 0 W W F2 = 2 530 N 3.90 m 𝐹2 = = 1467 N 1.40 m  F y = − F1 + F2 − W = 0 −𝐹1 + 1476 N − 530 N = 0 𝐹1 = 946 N Alternatively method of finding 𝐹2 Let fulcrum be the point or rotation Hence, ෍ 𝜏 = 𝐹1 ℓ2 − 𝑊(ℓ𝑊 −ℓ2 ) = 0 𝐹1 ℓ2 − 𝑊(ℓ𝑊 −ℓ2 ) = 0 𝐹1 × 1.4 − 530 3.90 − 1.4 = 0 530 × 2.5 𝐹1 = = 946𝑁 1.4 ❖Example 4 ❖An 80-kg man balances the boy on a teeter-totter as shown. Ignore the weight of the board, What is the approximate mass of the boy? ❖ Let force from the boy = 𝐹1 ❖ Let force from Fulcrum = 𝐹2 Let F = mg and g = 10 m/s2  = F  2 2 − W W = 0 W W F2 = 2 800 N 5m 𝐹2 = = 1000 N 4m ෍ 𝐹𝑦 = −𝐹1 + 𝐹2 − 𝑊 = 0 ≡ −𝐹1 + 1000 N − 800 N = 0 𝐹1 = 200 N F = mg 200 = m × 10 m = 20 kg Alternative method 𝑚𝑏𝑜𝑦 ℓ𝑏 − 𝑔𝑚𝑎𝑛 (ℓ𝑚 ) = 0 𝑚𝑏𝑜𝑦 × 4 − 80 × 1 = 0 80 × 1 𝑚𝑏𝑜𝑦 = = 20 𝑘𝑔 4 2 Assignment A hiker, who weighs 985 N, is strolling through the woods and crosses a small horizontal bridge. The bridge is uniform, weighs 3610 N, and rests on two concrete supports, one at each end. He stops one-fifth of the way along the bridge. What is the magnitude of the force that a concrete support ( ) exerts on the bridge (a) at the near end and (b) at the far end? zero F = 0. These two conditions will allow us to determine the magnitudes of F1 and F2. y F1 +y F2 Wh Axis + +x 1 5 L 1 2 L Wb Reference materials: 1. University Physics 2. Cutnel and Johnson 3. Giancoli 4. Physics for Scientists and Engineers