PHYSICAL SCIENCES 1 Concepts Lecture 5 Review Fall 017 1. Rotation axis: axis in which rigid body rotates about. It is perpendicular to the plane of rotation.. Angle θ: The angle at which the rigid body rotates through about the rotational axis. We typically use θ = 0 rad at the x-axis (i.e. θ is the standard angle). We define counterclockwise rotations from θ = 0 rad to be positive. For rotational motion, the units of θ are radians, where π rad = 360. 3. Angular displacement θ = θ f θ i : Counter-clockwise displacements are positive θ > 0. Clockwise displacements are negative θ < 0. 4. Arc-Length s: is the length of a circular arc. For a circular arc with radius r, the arc-length that spans an angular displacement θ is s = r θ. (1) 5. Angular Velocity: defined as the rate of angular displacement in time, ω = dθ dt Without Calculus: ω = θ t The sign of ω follows the same sign of θ. All points on a rigid rotating body share the same ω. 6. Tangential speed v t : defined as the magnitude of the velocity of an object (or a fixed point on the rigid object) along the curve. It is given by the rate of change of the arc-length s in time, v t = ds dt = rω Without Calculus: v t = rω (3) where r is the radial distance between the object (or fixed point) and the axis of rotation. While all points on a rotating rigid body share the same ω, their tangential speed will increase with its radial distance from the center. 7. Angular Acceleration α: defined as the rate of change of angular speed in time, α = dω dt Without Calculus: α = ω t The sign of α is positive for rotations speeding up in counter-clockwise direction and negative for rotations speeding up in the clockwise direction. 1 () (4)
8. Tangential acceleration a t : describes the acceleration of an object (or point on object) around the arc-length of the curve. A prime example is a person accelerating along a circular track. The relation between the tangential and angular acceleration an object (or fixed point on a rigid object) at constant r is, a t = dv t dt = r dω dt = rα Without Calculus: a t = rα (5) If the tangential acceleration is nonzero, the object (or fixed point) is not executing uniform circular motion. Instead, the object will be executing nonuniform circular motion and the tangential speed v t (t) will be changing in time. 9. Uniform Circular Motion: motion that requires for an object to move around a circle at fixed radial distance from the center r and constant tangential speed v t. In this case, the object has only centripetal acceleration, which only changes the direction of the object s velocity. a c = v t r = ω r. (6) 10. Centripetal a c Acceleration versus Tangential Acceleration a t : Any object (or fixed point on a rigid body) moving along a curve with radius of curvature r and a tangential speed v t experiences a centripetal acceleration, a c = v t r = ω r. (7) This holds whether the object is executing uniform circular motion or nonuniform circular motion. If the object is additionally accelerating along the curve, the object will have a tangential acceleration, a t. In the case of nonuniform circular motion, both the tangential speed v t, and thereby the centripetal acceleration, a c = v t (t) r, will change in time. The total acceleration a of an object (or fixed point) along a curve of radius r is, a = a c (inward) + a t (along curve) (8) The magnitude of the total acceleration is just, a = a c + a t (9) 11. Rotational Kinematics for Constant α: Angle θ Angular Velocity ω Relation of ω and θ θ(t) = θ 0 + ω 0 t + 1 αt ω(t) = ω 0 + αt ωf ω 0 = α θ
When using these solutions for ω and θ, it is important to use radians (not degrees), and to incorporate the appropriate sign conventions. For example, a disk that is rotationally decelerating in the clockwise direction is slowing down in the clockwise direction and hence, has a positive angular acceleration α > 0. In contrast, a disk that is rotationally accelerating in the clockwise direction has negative angular acceleration α < 0. 1. Moment of Inertia I: a physical measure of how the mass of the rotating body is distributed about its axis of rotation. It describes an object s resistance to rotational motion. For a point mass, the moment of inertia is where r is the distance of the mass from the rotational axis. I pm = mr (10) For a system of N point masses, the moment of inertia of the system is just the sum of inertias, I sys = j I j,pm = m 1 r 1 + m r + m 3 r 3 + + m N r N (11) where r 1, r,, r N is the distance between each mass and the axis of rotation. 13. Angular momentum L: a physical quantity defined as a product of a moment of inertia I and an angular velocity ω of an object, L = Iω. (1) It quantifies the amount of rotational motion of the object. 14. If a system is isolated from the environment, the total angular momentum of a system is conserved, L tot = 0 L i,tot = L f,tot (13) L tot t = 0 With calculus: dl tot dt = 0 (14) 15. Torque τ: is a physical measure of how much a force acting on an object causes that object to rotate. It is the rotational analog to force. In this class, we will only be working with the magnitude and sign 1 of torque. τ = F (r sin φ) = F r (16) τ = (F sin φ)r = F r (17) where r is called the moment arm and is perpendicular to F and F t is called the tangential force and is perpendicular to r. See figures below. 1 Torque is a vector quantity and mathematically defined as the cross-product of the position vector to the force relative to the pivot and force, τ = r F. The cross-product takes two vectors, r and F, and creates another vector that is perpendicular to both vectors and has a magnitude of rf sin φ. If r and F lie in the xy-plane, the torque will be in the ±ẑ. For this class, we just call the torque positive or negative. However, what we really mean is that torque points in either the positive ẑ direction or the ẑ direction. 3
The magnitude of torque is, τ = rf sin φ (15) where r = r is the distance between the pivot and point of force application, F is the force applied, and φ is the angle between r and F. One can intuitively think of the sin θ term in torque as a measure of how perpendicular r and F are to each other.the more perpendicular r and F are to each other, the larger the torque will be. Hence, we can alternative write the torque as, 16. Sign of Torque: τ > 0 for forces that lead to counter-clockwise rotations about the pivot. τ < 0 for forces that lead to clockwise rotations about the pivot The sign convention can be summarized with the help of the right-hand rule: 4
17. Rotational Analog for Newton s nd Law: A net torque on an object changes an angular momentum of a system: dl dt = τ net = τ j. (18) j In case the moment of inertia I of the object remains constant, it is equivalent to say that the net torque causes the object to rotationally accelerate, where α is the angular acceleration. 18. Summary of Angular VS Linear Quantities Iα = τ net, (19) Pure Translation (Fixed Direction) Pure Rotational (Fixed Axis) Position: x Angle: θ Velocity: v = dx dt Angular Velocity: ω = dθ dt Acceleration: a = dv dt Angular Acceleration: α = dω dt Mass: m Moment of Inertia: I Momentum: p Angular momentum: L dp Newton s nd Law: x dl dt = F net,x (ma x = F net,x ) Newton s nd Law: dt = τ net (Iα = τ net ) Kinetic Energy: K = 1 mv Kinetic Energy: K = 1 Iω 19. Conditions for Equilibrium: No net force acting on the body: Fnet = j F j = 0 No net torque acting on the body: τ net = j τ j = 0 These two conditions must hold for any choice of pivot in order for the object to be in equilibrium. Static equilibrium occurs when an object is at rest. Dynamic equilibrium occurs when an object moves at constant velocity and angular velocity. 5
0. Moment of Inertia of Rigid Objects: Rigid objects are made of a bunch of infinitesimal point masses, δm j. We can find the moment of inertia of the object by summing over the moment of inertia for all these tiny masses, I sys = j δm j r j (0) where r is the radial distance from the rotational axis. One can use calculus, to find the moment of inertia for various shapes shown in the table below. We can convert the sum into an integral, j δmjr j I sys = dm r. For this class, you will never need to do this. Always refer to the table. 6
Selected Problems The difficulty of each problem is rated on a scale of 1-5, with 1 being easy (novice level) and 5 being the most difficult (expert level). For context, Sapling problems generally have a difficulty level of 1-3, and offline HW problems generally have a difficulty level of 3-4. Bonus questions in lecture have a difficulty level of 4-5. 1. (3/5) A bead is given an initial velocity and then circles indefinitely around a frictionless vertical hoop. Only one of the vectors in the figure below is a possible acceleration vector at the given point. Which one? Circle all the correct choices below. (a) Point (a) (b) Point (b) (c) Point (c) (d) Point (d) (e) Point (e). (/5) Torque Square In the overhead view of the figure shown below, five forces of the same magnitude F act on a strange merrygo-round; it is a square that can rotate about point P, at midlength along one of the edges. Rank the forces according to the magnitude of the torque they create about point P, greatest first. Hint: It might be easier to use τ = F r = F r sin θ. 3. In the figure below, a disk, a hoop, and a solid sphere are made to spin about fixed central axes by means of strings wrapped around them, with the strings producing the same constant tangential force F on all three objects. The three objects have the same mass M and radius R, and they are initially stationary. The corresponding moment of inertias for each of the objects is, I disk = 1 MR, I hoop = MR, and I sphere = 5 MR. (a) (3/5) Rank the objects according to the magnitude of their angular acceleration α hoop, α sphere, and α disk about their central axes. (b) (/5) Rank their angular speeds ω hoop, ω disk, and ω sphere, greatest first, when the strings have been pulled for a certain time t. Hint: What can we say about ω(t) when the angular acceleration is constant? 3 Note that some of the Sapling problems have a difficulty level of 3-4. 7
4. (3/5) Rank in order, from largest to smallest, the angular accelerations α a to α e of the rigid objects in cases (a)-(e) shown below. 5. The flywheel of a steam engine turns counter-clockwise with a constant angular velocity of 5π rad/s. When steam is shut off, the friction of the bearings and of the air stops the wheel in 700 s. (a) (3/5) What is the constant angular acceleration of the wheel during the slowdown? (b) How many revolutions does the wheel make before stopping? (c) (3/5) At the instant the flywheel is turning at.5 π rad/s, what is the tangential component of the acceleration of a point on the flywheel that is 50 cm from the axis of rotation? (d) (3/5) What is the magnitude of the net linear acceleration a of the particle in (c)? (e) (3/5) The flywheel has a total mass of M = 100 kg and is a solid cylinder with radius R = 1 m, Calculate the net work done by friction on the flywheel. 6. Starting from rest, a race car accelerates along a circular racetrack with radius r at constant tangential acceleration a t. (a) (/5) Write an expression for the tangential speed v t of the car as a function of time. (b) (3/5) Write an expression for the magnitude of the centripetal acceleration as a function of time, a c = a c = v t r. (c) (3/5) Find the magnitude of the car s total acceleration a net. (d) (5/5) If the coefficient of static friction between the road and the car s tires is µ s, at what time will the car start to slip? 8
7. (4/5) You pull on a roll of toilet paper with a given force F and observe the angular acceleration α. A few days later when the radius is half of what it was, you pull with the same force F and angle. (For simplicity, ignore the hollow tube; assume that the toilet paper goes all the way down to zero radius.) The ratio of the new α to the old is... (a) 1 (b) (c) 4 (d) 8 (e) 16 8. (/5) The figure below shows three situations in which the same horizontal rod is supported by a hinge on a wall at one end and a cord at its other end. Without many written calculation, rank the three scenarios according to the magnitudes of the specified forces, greatest first. (a) the force on the rod from the cord (b) the vertical force on the rod from the hinge (c) the horizontal force on the rod from the hinge 9. (3/5) The figure below shows a simplified model of the force needed to do a push-up. We assume that a persons body is a bar of mass M with a moment of inertia about the pivot (at the toes) of I T. The center of mass is a distance L from the pivot and that the arm force, F arm is purely vertical and applied at a distance of 4L/3 from the pivot. The height h is given by the person s arm extension and varies throughout the entire push-up. 9
(a) Find the value of the applied force arm force, F arm, needed for the object to be in equilibrium. Show that it is independent of the angle θ and thus, independent of the height, h. (b) What is the value of the normal force F N exerted on the person by the ground at the person s toes? 10. (4/5) Stick on A Corner: A stick with mass m and length l leans against a frictionless wall, with a quarter of its length hanging over a corner, as shown in the figure. It makes an angle θ with the horizontal. Assuming that there is sufficient friction at the corner to keep the stick at rest, what is the total force that the corner exerts on the stick? Hint: We suggest that you break the total corner force into x and y components F c = (F cx, F cy ). 11. (4/5) Tip or Slip? A uniform rectangular wood block of mass M, with length b and height a, rests on an incline as shown. The incline and the wood block have a coefficient of static friction, µ s. The incline is moved upwards from an angle of zero through an angle θ. At some critical angle the block will either tip over or slip down the plane. Determine the relationship between a,b, and µ s such that the block will tip over (and not slip) at the critical angle. The box is rectangular, and a b. 1. (5/5) Balancing Disk Ultimate Challenge: A stick with mass m and length R is pivoted at one end on a vertical wall. It is held horizontal, and a disk with mass m and radius R is placed beneath it, in contact with both it and the wall, as shown right. The coefficient of friction between the disk and the wall is µ w, and the coefficient of friction between the disk and the stick is µ s. If the objects are released, what are the minimum values of µ w and µ s for which the system doesn t fall? 10
3 Solutions to Selected Problems 1. Answer: Point (d) The only two forces acting on the bead are the normal force, which points radially inward, and gravity F g = mgŷ, which always points downward. Summing these two vectors up at each point gives the net acceleration of the bead. Point (d) is the only place where the drawn acceleration arrow is possible. Remark: Since the radial a r = v r component always points radially inward, the acceleration vector in any arbitrary circular motion can never have a radially outward component. This immediately rules out choices (a) and (c).. Answer:τ 5 > τ 4 > τ > τ 1 > τ 3 The figure right shows the corresponding position vectors that point from the pivot to where each force is applied. The side-length of the square is d. Since forces F 1, F, and F 5 lie along the x and y axes, the best way to calculate the torque from each of these forces is to use, τ = F r. Doing so gives, where F 1 = F = F 3 = F. τ 1 = F (0) = 0 (1) ( d τ 4 = F = ) F d () τ 5 = F d (3) The position vector of forces F 1 and F is the same and points in the ˆx direction. Thus, the best way to calculate torque is τ = F r, where F = F sin θ and r = d. While we don t know the angle θ between these two forces and the position vector, we can see that F 1 is almost parallel to r (i.e. arrow in orange), and thus, F has a larger perpendicular component than F 1. This means that, d τ 1 = F 1, = (F sin θ 1)d < τ (4) d τ = F, = (F sin θ )d < τ 4 (5) where sin θ 1 < sin θ 1. Putting this together gives, τ 5 > τ 4 > τ > τ 1 > τ 3. 3. Disk, Hoop, and Sphere (a) Answer: α sphere > α disk > α hoop According to Newton s nd Law for rotational motion, τ net = Iα α = τ net I (6) 11
(b) Since the torque on each of the three rigid objects is the same, τ net = F R, the corresponding angular accelerations are, α disk = F R I disk = F R mr = F R mr α hoop = F R = F R I hoop mr = F R mr α sphere = F R = F R I mr sphere 5 (7) (8) = 5F R mr. (9) This gives us, α sphere > α disk > α hoop. We should expect this because objects with mass closer to their center of mass, such as the sphere, are easier to rotate than objects with mass further from their center of mass, such as the hoop. Also note how the torque, and thereby the angular acceleration, is positive since the force causes counter-clockwise rotations. Answer: ω sphere > ω disk > ω hoop Since the angular acceleration for each of the three objects is constant, we can use the kinematic equations for constant α to find ω after time t. Since each object starts from rest, ω 0 = 0 for all three objects. This gives us, Therefore, ω sphere > ω disk > ω hoop. F R t ω disk = α disk t = mr ω hoop = α hoop t = F R t mr 5F R t ω sphere = α hoop t = mr (30) (31) (3) 4. Answer: α b > α a > α c = α d = α e Each of the five cases consists of two point masses connected by a massless rod and subject to a constant force F. According to Newton s nd Law for Rotational Motion, the angular acceleration is, α = τ net I sys. (33) The moment of inertia for each system is, I sys,j = m j rj, where j = {a, b, c, d, e} and r j is the distance from the center of the rod to each point mass. The net torque for each system is τ net = F,j r j, where F perp,j is the force perpendicular to the rod. Putting these two together gives us a general expression for the angular acceleration of the system, α j = τ net I sys = F,j m j r j (34) 1
Using equation 34 to calculate the angular acceleration in each case gives, 5. Rotating Flywheel α a = F,a m a r a = α b = F,b m b r b = α c = F,c m c r c = α d = F,d m d r d = α e = F,e m e r e = 1 N ( kg)(1 m) = 1 rad/s (35) N ( kg)(1 m) = 1 rad/s (36) 1 N (4 kg)(1 m) = 1 4 rad/s (37) 1 N ( kg)( m) = 1 4 rad/s (38) 1 N ( kg)( m) = 1 4 rad/s (39) (a) Answer: α = π 1440 rad/s = 7 10 4 rad/s In this problem, we are told the initial angular velocity, ω 0 = 5π rad/s, the final angular velocity, ω f = 0 rad/s, and the time for the friction of the bearings to bring the wheel to rest, t f = 700 s. Since the wheel undergoes constant angular acceleration α, we can use the kinematic equations for angular velocity to solve for α. ω f = ω 0 + αt f α = ω f ω 0 t f 5π rad/s = = π 700 s 1440 rad/s (40) (b) It is important to note that the negative sign on the angular acceleration describes the wheel slowing down in the counter-clockwise direction. Answer: 9000 revolutions Now that we know the angular acceleration of the wheel, we can calculate the number of revolutions θ the wheel rotates through before coming to a stop. There are two equations in which we can use to solve for θ, ω f ω i = α θ (41) θ = ω 0 t f + 1 αt f (4) While either will work, the author of this guide personally prefers equation 41, so we will go with this one. Solving for θ from equation 41 gives, θ = ω f ω i α = 5π rad /s π 1440 rad/s = 18000π rad (43) Since the question asked for revolutions as opposed to radians, we need to convert from radians to revolutions. For every revolution, there is π radians. Using this conversion factor, we can solve for how many revolutions the wheel makes, θ = 18000π rad 13 ( 1 rev ) = 9000 rev (44) π rad
(c) Answer: a t = π 880 m/s = 3.5 10 4 m/s Although we are given ω =.5π rad/s, the tangential acceleration of a point on the flywheel only depends on α and the radial distance from the center r, a t = αr (45) Given that the radial distance of the point from the center is r = 1 m, the corresponding tangential acceleration is, a t = π 1440 rad/s( 1 ) m = π 880 m/s = 3.5 10 4 m/s (46) Note how the rad unit disappeared in our final expression for a t because it isn t a true unit such as meters or seconds. The negative sign indicates the tangential acceleration points against the curve and that the point is decelerating along the tangential direction. [ (d) Answer: a = 5π 8 m/s ] [ (inward) π 880 m/s] (along) The total acceleration of the point on the wheel has both a centripetal and radial component, 4 a = ac (inward) + at (along) (47) The centripetal acceleration changes quadratically with ω, ( 5 ) ( 1 ) [ 5π a c = ω r = π rad/s m = m/s ] (48) 8 With this and the tangential acceleration from part (c), we can find the total acceleration of the point at this time, [ 5π a = m/s ] [ π (inward) 8 880 m/s] (along) (49) (e) Answer: W friction = (5π) kg m /s = 6168.5 J According to the Work-Kinetic energy theorem, W friction = K = 1 I wheel ( ) ωf ω 0 (50) where I wheel is the moment of inertia of the wheel. Since the wheel is a solid cylinder with mass M = 100 kg and R = 1 m, its moment of inertia is, I wheel = 1 MR = 50 kg m (51) The initial and final angular speeds are, ω 0 = 5π rad/s and ω f = 0 rad/s, respectively. Substituting these values into equation 50 gives, W friction = 1 ( ) I wheel ωf ω 0 = 1 (50 kg m )( 5π rad /s ) = (5π) kg m /s (5) ( ) 4 Expert Note: One can more efficiently describe the inward radial direction by ˆr = cos θˆx + sin θŷ, which points inward for all points along the unit circle, and the tangential (along) direction by ˆθ = sin θˆx + cos θŷ. For this class, you will not need to use polar coordinates axes.{ˆr, ˆθ, ẑ}, to describe the net acceleration. 14
6. Race Car (a) Answer: v t = a t t Since the car has constant tangential acceleration and starts from rest (i.e. v 0 = 0), the tangential velocity will be, v t = v 0 + a t t = a t t (53) (b) Answer: a c (t) = (att) r As the car accelerates along the track with radius r, it has a tangential speed v t = a t t, which increases in time. The centripetal acceleration at every point in time is, a c = v t r = (a tt). (54) r From this expression, we see that the centripetal acceleration increases quadratically with time. (a (c) Answer: a = a = tt) 4 + a r t The total acceleration of the car has a centripetal component, which points inward, and tangential component, which points along the circle. Therefore, a = (att) r (inward)+a t (along). The magnitude of the total acceleration is just the square root of the sum of these two components squared, [(at a = a = a c + a t = t) ] + a r t (55) (d) Answer: t = 1 at (r [(µ s g) at ]) 1/4 The free-body diagrams from the car from the top and side-view are shown below. The only force responsible for accelerating the car is the static friction force between the tires and the road, F s µ s F N. The constraints on the motion are that the car accelerates in the xy plane and does not accelerate in the z-direction (i.e. does not levitate off the race track). This means there is no net force acting in the z-direction, such F net,z = ma z = 0, 15
and the normal force on the car is the same magnitude as gravity, F N = mg. According to Newton s nd Law, F net = F S = m a (56) The force of static friction is parallel to the total acceleration, and hence, has a changing magnitude and direction in time. 5 We know that the magnitude of static friction force is bounded by, F s µ s mg. We can use this to find an upper bound on the acceleration a = a = a c + a t. Taking the magnitude of Newton s nd law give, [(at F net = F t) s µ s mg = ma µ s g = 4 ] r + a t (57) It is important to notice that the maximum possible value of the car s acceleration is independent of mass and only depends on g and the coefficient of static friction for your tires µ s 1 for a dry road (or µ s 0. for a wet road). We can now solve for the time t for the car to slip using equation 57, a t + (a tt) 4 r = (µ s g) (a t t) 4 r = (µ s g) a t (a t t) 4 = r [ (µ s g) a t t = 1 ( r [ (µ s g) a t at ] ]) 1/4 (58) ( ) Unitwise this makes sense because we have [t] = s m 4 1/4 m s = s. However, let s substitute 4 in a few numbers to see if our result is reasonable. If the car has a typical acceleration of a t = 4 m/s, is on a dry road with µ s = 1, and rounding a race car track bend with r = 00 m, the time for it to slip would be t = 10.58 s. The car s corresponding tangential velocity would be v t = a t t = 93.1 mph. This seems reasonable. How does this compare with if the maximum speed the driver decided to not accelerate around the curve and instead travel with a constant speed? At constant speed, the acceleration only has a inward centripetal component, a c = v max r = µ s g v max = µ s gr (59) Given the same specs as before, the driver would be able to do so safely at a slightly higher speed v max = 97.4 mph. The moral of the story is that it is slightly safer to coast around a curve than to accelerate. It is even safer to drive on a banked (inclined) road, but we will leave this up to the student to read about! 5 This problem contrasts to your Offline HW problem with Javier riding on the Merry-go-Round at constant tangential speed. In this case, the acceleration only had a centripetal component (inward) since Javier was not accelerating along the curve. 16
7. Answer: (d) The acceleration α is caused by the applied torque by τ = I α, so α = rf mr. For our toilet paper roll, the mass depends on the radius, since m = ρ Volume r. This means that α rf r 4 1 r 3 so if r is halved the angular acceleration will go up by a factor of 8. 8. Rod attached to ceiling (a) Answer: F T,1 = F T,3 > F T, The free-body diagram for the three rods is shown below, where F H is the hinge force 6, F g is the gravitational force, and F T is the tension force. The perpendicular component of the tension force to the rode is F T, = F T cos θ, where θ = 50 for cases (1) and (3) and θ = 0 for case (). We aren t given any specific details of the system, but we will assume the rod is of length L. In all three cases, gravity F g acts downward on the rod at its center of mass, L. Since the rod is in static equilibrium, we have τ net = 0 (60) F net,x = 0 (61) F net,y = 0 (6) We can relate the force of gravity to the tension force by calculating the net torque about the hinge, τ n et = τ T τ g = 0 τ T = τ g F T, L = F gl F T = F gl cos θ With this expression, we find that F T,1 = F T,3 > F T, since cos 50 < 1. (63) wall. 6 The hinge force is simply the normal force that the hinge exerts on the rod because the rod cannot penetrate the 17
(b) Answer: F H,y,1 = F H,y, = F H,y,3 The vertical component of the hinge force F h,y is simply the perpendicular component of the force to the rod. This means that we can relate F h,y to F g by finding the net torque about the end of the rod, 7 τ net = τ g τ H = 0 τ H = τ g F H,y L = F gl F H,y = F g This expression tells us that the vertical component of the hinge force is the same in all three cases. Remark: It is important to note that we could have come to the same conclusion using equation 6, (64) F net,y = F H,y F g + F T, = 0 F H,y = F g F T, (65) (c) However, we would need to solve for F T, = F T cos θ, as we did in part (a). Answer: F H,x,1 = F H,x,3 > F H,x, We can solve for the X-component of the hinge force F Hx using equation 61. Before jumping into the calculation, though, it is important to notice the x-component of the hinge force to be zero in case because both the tension and gravitational forces point in the ±ŷ-direction. Hence, the rod is not being pushed into the wall in the horizontal direction and therefore, F H,x, = 0. Now for the other two hinge forces, F net,x,1 = j F j,x = F H,x,1 F T,x,1 = 0 F H,x,1 = F T sin θ (66) F net,x,3 = j F j,x = F H,x,3 + F T,x, = 0 F H,x,3 = F T sin θ (67) 9. Push-up This gives us F H,x,1 = F H,x,3 > F H,x,. (a) Answer: F = 3 4 mg The free-body diagram for the person is shown right. In this problem, there are two unknowns: the normal force acting on the person s toes, F N, and the arm force, Fa rm = F ŷ. Since the person is in static equilibrium, we have the following constraints, F net,x = 0 (68) F net,y = 0 (69) τ net = 0 (70) 7 For static equilibrium, the net torque about any pivot must be zero. 18
(b) For the first part of the problem, we just need to calculate the arm force F arm = F ŷ and show that it is the same for any arm length h (i.e. is independent of θ). Since we don t know the normal force, we can place the pivot the person s toes to ensure that the torque due to the normal force is zero (i.e. τ N = 0 since the moment arm is zero). The constraint of the net torque then gives us, τ net = τ arm τ g = 0 τ arm = τ g F r arm, = F g r g, 4L cos θ F 3 = mgl cos θ F = 3 mg (71) 4 Since our expression does not depend on θ, the required arm force in a plank is independent of arm length, h. Answer: F N = 1 4 mg We can calculate the normal force acting on the person s toes using the constraint of the net force in the y-direction, 10. Answer: Fc = ( mg 3 tan θ, mg ) F net,y = F N + F mg = 0 F N = mg F = mg 4 (7) The free-body diagram is shown for the stick. In order for the stick to be in static equilibrium, τ net = 0 (73) F net,x = 0 (74) F net,y = 0 (75) Note that gravity acts on the center of mass, which is a distance l 4 from the corner. In this problem, there are three unknowns: the x and y components of the corner force, (F Cx, F Cy ), and the magnitude of the normal force of the wall on the stick F N,W. This means that we will need to use all three equations to solve for our unknowns. Since we do not know either components of the corner forces, we should place our pivot at the corner and calculate the torque this point. The reason is that the torque by the corner force at this point is zero because the moment arm is zero. According to equation 73, we have τ net = τ N,W τ g = 0 τ N,W = τ g F N,W r W, = mgr g, 3l F N,W 4 sin θ = mg l 4 cos θ mg F N,W = 3 tan θ (76) 19
We can relate the two horizontal forces F Cx and F N,W using equation 74, We can solve for F Cy using equation 75, F net,x = F Cx F N,W = 0 F Cx = F N,w = mg 3 tan θ (77) F net,y = F Cy F g = 0 F Cy = F g = mg (78) This gives us F ( ) C = mg 3 tan θ, mg 11. Answer: µ s > b a The free-body diagram of the box is shown below. When the box is just about to tip, imagine the left surface of the box is ɛ away from the incline, such that the normal force is only at the lower right edge of the box (i.e. at the green dot). Consequently, the static friction force acts at the lower right edge of the box. Newton s nd Law for the box along x and y direction reads, F x,net = F g sin θ F s = 0 (79) F x,net = F N F g cos θ = 0 (80) where a x = 0 and a y = 0 because the box is not slipping. In order for the box not to slip, F s = F g sin θ µ s F N = µ s F g cos θ µ s > tan θ (81) The net torque about the green dot is, ( ) a ( ) b τ net = F g sin θ F g cos θ (8) According to the rotational analog of Newton s nd law, in order for the box to tip ( ) a ( ) b τ net = Iα > 0 F g sin θ > F g cos θ tan θ > b a (83) Putting equations 81 and 83 gives us, µ s > b a (84) Since µ s < 1, this relation tells us that the box will only tip if b < a (i.e. the box is taller than it is wide). 0
1. Answer µ w > 1 and µ s > The free-body diagram for the disk and the stick (which has mass m) is shown below. In this problem, there are 6 unknown variables: the normal force and static friction force from the wall, F N,W and F W, the normal force and static friction force with the stick, F N,S 8 and F S, and the coefficient of static friction at each surface, µ s and µ W. The forces laws for friction provide two constraints, F S µ S F N,S (85) F W µ W F N,W (86) Both the stick and disk are in static equilibrium, and the following conditions hold for each object separately, F net,x = 0 (87) F net,y = 0 (88) τ net = 0 (89) We can directly solve for the normal force on the disk from the stick, F N,S using Newton s nd Law for the stick, F net,y,stick = F N,S mg = 0 F N,S = mg (90) By Newton s 3rd Law, the normal force exerted by the disk on the stick is equal and opposite in direction to the normal force exerted on the disk from the stick. Since we now know all but a single vertical forces acting on the disk, We can solve for F W using Newton s nd Law for the disk in the y-direction, F net,y,disk = F W F g F N,S = 0 F W = F g + F N,S = mg (91) We can now relate the static friction force from the stick F S and from the wall F W by calculating the net torque about the disk s center of mass. At this point, only the torques due to F S and F W will be nonzero, τ net = τ S τ W = 0 τ S = τ W F S R = F W R F S = F W = mg (9) 8 For each normal and static friction force on the disk, there is an equal in magnitude, but opposite in direction force acting on the wall/stick. 1
Since we know both F N,S and F S, we can find the conditions for µ s (eqn. 85), F S µ s F N,S mg µ s mg µ S (93) We now know all forces in the horizontal direction except for F N,W. Therefore, We can solve for F N,W using Newton s nd Law for the disk in the x-direction, F net,x = F N,W F S = 0 F N,W = F S = mg (94) Since we know both F N,W and F W, we can find the conditions for µ W (eqn. 86), F W µ W F N,W mg µ W (mg) µ W 1 (95)