Sp08Exam2Review - Review Exam 2 Chapters 5 - 8 Newton's...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Review Exam 2 Chapters 5 - 8 Newton's Second Law n The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass a = SF/ m Algebraically we can also write, SF = m a More About Newton's Second Law n SF is the net force n This is the vector sum of all the forces acting on the object n Newton's Second Law can be expressed in terms of components: n n n SFx = m ax SFy = m ay SFz = m az Dynamic Equilibrium: Ex. 2 A car with a weight of 15,000N is being towed up a 20 slope at constant velocity. Friction is negligible. The tow rope is rated at 6000N. Will it break? Know: q = 20, w = 15,000N. Find: T Ex. 2 Dynamic Equilibrium: Tilted Coordinate System Apply Newton's 2nd Law (a=0) r F = 0 x F = T - w sin q = 0 F = n - w cos q = 0 y From the x equation T = w sin q = (15,000N) sin 20 0 = 5130N Since T < 6,000N the rope won't break. Note, y equat. not needed. Check solution at q = 0 and 90 Ex.2:Dynamic Equilibrium: Un-tilted Coordinate System Fx = 0 T cos 20o - n sin 20o = 0 Fy = 0 T sin 20o + n cos 20o - mg = 0 cos 20 o From the x equation : n = T sin 20 o Sub. in y equation cos 20 o T sin 20 + T cos 20 o - mg = 0 sin 20 o o T (sin 2 20 o + cos2 20 o ) - mg sin 20 o = 0 T = mg sin 20 o Weight & Apparent Weight Weight is the force exerted by the earth on a mass m. w = mg, where g = 9.80m/s2. Consider a scale and weight at rest. When the weight comes to rest with respect to the spring Fsp = w The pointer is calibrated to read Fsp which is equal to w at equilibrium (a = 0) Suppose a 0 Weight & Apparent Weight Consider a fish hanging from a scale in an elevator moving up with acceleration a. The y component of N's 2nd Law is T=Fsp F = F - mg = ma y sp \ Fsp = m(g + a) The spring dial comes to rest with the force Fsp=m(g+a) > w m(g+a) is the apparent weight Weight & Apparent Weight n Now consider the elevator accelerating downward. F = F - mg = m(-a) y sp \ Fsp = m(g - a) n n T=Fsp Now the scale reads Fsp= m(g-a) <w If a=g, called freefall, the apparent weight is 0 Friction n Static friction force: n variable 0 < F fs m sn constant n Kinetic friction force: n F fk = m k n < m sn Friction Forces Forces of Friction n n The direction of the frictional force is opposite the direction of motion, or intended motion, and parallel to the surfaces in contact The coefficients of friction are nearly independent of the area of contact Newton's Third Law n If two objects interact, the force F12 exerted by object 1 on object 2 is equal in magnitude and opposite in direction to a force F21 exerted by object 2 on object 1 n F12 = - F21 n Note on notation: FAB is the force exerted by A on B n Note: read FAB as Force of A on B Interacting Masses A 10.0 N force pushes on a 4.00 kg box (m1) which is touching a 2 kg box( m2). What force does the 2.00 kg box experience and what is its acceleration? Apply Newton's 2nd law to each box noting that there are reaction forces between the boxes. Interacting Boxes F - F12 = m1 a F21 = m 2 a The a's are the same because the boxes are in contact. Add the equations and make use of the action-reaction force concept: F12 = -F21. 10.0N F = (m1 + m 2 )a fi a = = 1.67m / s 2 6.00kg F21 = m 2 a = (2.00kg)1.67m / s = 3.33N 2 Normal Forces The normal force (force of the table on monitor) is the reaction of the force the monitor exerts on the table (W = mg) n Normal means perpendicular to the surface. Free Body Diagram n n In a free body diagram, you want the forces acting on a particular object The normal force and the force of gravity are the forces that act on the monitor What is the Normal Force on Each Box and all of the Boxes on the Table? Consider all the Boxes as One Box with m = m1+m2 + m3 Use a free body force diagram & apply Newton's 2nd Law. Fy = 0 ntableonbox - (m1 + m2 + m 3 )g = 0 ntableonbox = (m1 + m2 + m 3 )g Consider Each Box Separately Make a free body force diagram for each box and a[ply Newton's 2nd Law to each. Fy = 0 n2on1 - m1g = 0 \ n2on1 = m1g Fy = 0 n3on2 - m2 g - n1on2 = 0 n1on2is the reaction pair to n2on1 n1on2 = -n2on1 = m1g \ n3on2 = m2 g + m1g Diagram for m3 Fy = 0 ntableon3 - m 3 g - n2on 3 = 0 n2on 3is the reaction pair to n3on2 n2on 3 = -n3on2 = (m1 + m2 )g \ ntableon3 = (m1 + m2 + m 3 )g Multiple Objects n If we are considering a problem in which friction between objects is important it will be necessary to take into account the normal force and the reaction normal force to compute the frictional force, i.e. f = n. Motion of Multiple Boxes w/friction The coefficient of kinetic friction between the boxes and between the lower box and the floor is 0.3 What is the acceleration of the lower box? Motion of Multiple Boxes w/friction F1x = f1 -T1 = m1a1 = -m1a (a1 = -a) F1y = n1 - m1g = 0 fi n1 = m1g F2 x = Tpull - f1 - f2 -T2 = m2 a2 = m2 a F2 y = n2 - n1 - m2 g = 0 n1 = n1 = m1g n2 = n1 + m2 g = (m1 + m2 )g Motion of Multiple Boxes w/friction f1 = f1 = m k n1 = m k m1g T1 = T2 = T f2 = m k n2 = m k (m1 + m2 )g F1x =mk m1g -T = -m1a fi T = m k m1g + m1a F2 x =Tpull -T - m k m1g - m k (m1 + m2 )g = m2 a Substitute for T in the F2 x equation Tpull - m k (3m1 + m2 )g = (m1 + m2 )a a= Tpull - m k (3m1 + m2 )g (m1 + m2 ) = 1.77m / s 2 Uniform Circular Motion The acceleration of uniform circular motion points to the center of the circle. Thus the acceleration vector has only a radial component ar. This acceleration is conveniently written in the rtz-coordinate system as r = inward radial positive t = tangential ccw is positive z = upward positive Dynamics of Uniform Circular Motion The usefulness of the rtzcoordinate system becomes apparent when we write Newton's second law in terms of the r-, t-, and z-components, as follows: Turning the corner I Turning the corner I Turning the corner I Turning the corner I Turning the corner I The only difference from the example of the ball and string is that the tension force in the string is replaced by a static friction force toward the center. From the radial equation the speed is rf s v= m The maximum allowed speed for UCM will be when the frictional force is a maximum. Frictional Forces Turning the corner I The maximum static frictional force is when From the z component of N's 2nd law f s = m sn F = n - mg = 0 f = m mg z s s fi n = mg v max rm s mg = = m s rg m m s = 1 (Table 6.1) v max = (1.0)(50.m)(9.80m / s 2 ) = 22m / s Water Pail A pail of water is rotated in a vertical circle of radius 1.00 m. The pail undergoes UCM. The water also undergoes UCM if the water is in contact with the bottom of the pail. This means the pail must exert a normal force on the water. Let us assume the water is going in UCM and use Newton's second law equations to analyze the motion of the water. Water Pail For the water: Fr = n + mg sin q = mar v2 For circular motion a r = r v2 Fr = n + mg sin q = m r For circular motion the water experiences a normal force with magnitude: v n = m - mg sin q r v n = mg( - sin q ) gr 2 2 Water Pail v2 n = mg( - sin q ) gr v2 (rw )2 w2 n = mg( - sin q ) = mg( - sin q ) = mg( - sin q ) gr gr g /r g Let us define w c = r w2 n = mg( 2 - sin q ) wc w2 at q = 90 n = mg( 2 -1) wc Water Pail Imagine swinging a bucket of water over your head. If you swing the bucket quickly, the water stays in. But you'll get a shower if you swing too slowly. In order for the water to move in UCM there must be a normal force pushing on the water. The critical angular velocity wc is that at which the normal force disappears Why Does the Water Stay in the Bucket? w>wc Why Does the Water Stay in the Bucket? wc w=wc w=wc Why Does the Water Stay in the Bucket? Nonuniform Circular Motion Nonuniform Circular Motion The force component (Fnet)r creates a centripetal acceleration and causes the particle to change directions. The component (Fnet)t creates a tangential acceleration and causes the particle to change speed. Force and acceleration are related to each other through Newton's second law. Example: Pendulum A 2 kg ball is tied to a string which is 0.500 m long. The ball swings in a vertical circle under the influence of gravity. When the ball makes an angle of 20 with the vertical its speed is 1.50 m/s. a) What is at? b) What is ar? c) What is the tension in the string? d) What is the total acceleration? Pendulum - aq n What is the tangential acceleration? n Note that since we measure angles from the positive x axis q is increasing, from 250. Hence aq is positive. r r Ft = mg sin qt^ = mat r at = +g sin qt^ n r at = 9.80m / s 2 sin(20) = +3.35m / s 2 At this time the tangential speed is increasing. Example: Pendulum n What is the centripetal acceleration? r v 2 (1.5m / s)2 s t = ^ ar = = 4.5m / s r r 0.5m n Fr = mar What is the tension in the string? v2 T - mg cos q = m r v2 T = m(g cos q + ) = 2.00kg[(9.80m / s 2 ) cos 20 + 4.5m / s 2 ] = 13.7N r Example: Pendulum Pendulum - at What is at when the ball passes through the vertical? r r Ft = -g sin qt^ = mat r fi at = -g sin qt^ ...
View Full Document

Ask a homework question - tutors are online