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1ENGINEERING MECHANICS CHAPTER 7 School of EngineeringENGINEERING MECHANICSCHAPTER 7Kinetics of Linear Motion –Newton’s 2nd Law MethodClick on VIEW and select SLIDE SHOW to view the presentation.
27.1 BriefKinetics is the study of motion with the forcesassociated with it. In the earlier chapter underKinematics, we have studied motion withoutconsidering the forces. For analysis of body withforces, Newton’s three laws of motion are applicable.Newton’s First Law:Every body remains at rest or maintains a constantvelocity in a straight line unless an unbalanced forceacts upon it..
3Newton’s Second Law:A body acted upon by an unbalanced force experiencesan acceleration that has the same direction as theforce and a magnitude that is directly proportional tothe force.Newton’s Third Law:For every force, there is an equal and opposite reaction.
47.2 Newton’s Second Law:Newton stated that acceleration is proportional to the applied force with the mass as the constant of proportionality.Newton’s second law can be written in mathematical form as:F = ma for a single forceF = ma for several forcesWhere F is the force (N)m is the mass (kg)a is the acceleration (m/s2)
57.3 Free body diagramWe note that since the motion is caused by forces acting on the body, the first step would be to determine all the forces acting on it. This is set out in the form of a free body diagram indicating clearly all external and internal forces.The solution involves summing all the forces to obtain the force that causes unbalance and equating that with mass times acceleration i.e. F = ma.7.4 Sign conventionWe choose the direction of the resulting motion as thepositive direction (note that this is different from Statics)However, any direction can be selected as positive butthis must be strictly adhered to throughout the analysis.
6Uniform motionA motion is called uniform motion when theacceleration is constant.The three equations of motion with constant acceleration are:v = vi + atv2 = vi2 + 2ass = vit + ½at2where v is the final velocity (m/s)vi is the initial velocity (m/s)a is the acceleration (m/s2)s is the displacement (m)t is the time (s)
7a horizontal force of 10 N, towards the right, Example 7.1A box of mass 5 kg rests on a frictionless horizontal surface. Determine the normal reaction at the surface and the acceleration of the box when it is acted on by,a horizontal force of 10 N, towards the right,a force of 10 N at 60 with the horizontal (downwards and leftwards),combination of forces in forces in (i) and (ii).Solution:i) For x-direction, Fx=10,since F = ma10 = 5 x aa = 2 m/s2 Fy = 0, (no motion upwards)Nr – 5g = 0Nr = 49.1 N Nr105g ( g = 9.81 m/s2 )F.B.D
8ii) Along the x-direction, Fx = - 10cos60 = - 5 N F = ma Nr605g10ii) Along the x-direction,Fx = - 10cos60= - 5 NF = ma- 5 = 5 x aa = - 1 m/s2 aii)Fy = 0, (no motion upwards)Nr - 5g – 10sin 60 = 0Nr = N
9iii) Along the x-direction, Fx = 10 - 10 cos60 = 5 N F = ma 5 = 5 x aa = 1 m/s2 Fy = 0, (no motion upwards)Nr - 5g – 10sin 60 = 0Nr = N Nr605g10a
10In the x-direction, the unbalanced force on the car is Example 7.2A car of mass 950 kg accelerates up a slope with the engine pulling with a force of 1500 N. If the acceleration is 0.25 m/s2, determine the angle of the slope.a = 0.25 m/s21500 N950g NNrSolution:In the x-direction, the unbalanced force on the car isFx = 1500 –950g sinF = ma1500 –(950 x 9.81 x sin) = 950 x 0.25Hence, = (Ans.)
11The net force is Fx = T – Fr – (500,000 x g x sin) Example 7.3A 500 tonne train is being hauled by a locomotive up an incline of 1 in 80. Its velocity decreased from 45 km/h to 15 km/h in 50 s. If the resistance to the motion is 65 N/tonne, calculate the average tractive force T, applied by the locomotive.x801Fr = 65 x 500 NT(500 x 1000x g)NrSolution:The net force is Fx = T – Fr – (500,000 x g x sin)Since sin = 1/80, Fx = T – 93,812.5F = maT – 93,812.5 = 500,000a (1)
12Substitute into equation (1), T – 93,812.5 = 500,000a x801Fr = 65 x 500 NT(500 x 1000x g)NrThe acceleration of the train, in the x-direction can be obtained from:v = u + at15,000 / 3600 = / a x 50a = m/sSubstitute into equation (1),T – 93,812.5 = 500,000aT – 93,812.5 = 500,000 x (-0.167)T = 10,312.5 N
13Two blocks, A and B of masses 10 kg and 6 kg Example 7.4Two blocks, A and B of masses 10 kg and 6 kgrespectively, are connected by an inextensible stringover a light frictionless pulley as shown. The systemis released from rest. Determine the acceleration ofthe mass A if the table is smooth.B6 kgA10 kg
14NrT10g6gABaSolution:The two masses are connected by inextensible cord, hence their accelerations are both equal.For block A, choose positive x direction towards the right (along the direction of acceleration).Fx = TF = maT = 10a (1)
15Substitute (1) into (2), T = 10a ------ (1) NrT10g6gABaFor block B, choose positive y direction downward (along the direction of acceleration).Fy = 6g - TF = ma6g - T = 6a (2)Substitute (1) into (2), T = 10a (1)6g – 10a = 6aa = m/s2
167.5 FrictionWhen the body moves on a rough surface, there will be friction resisting the motion.The friction force equation is (Refer Chapter 5)F = kNwhere F is the frictional force (in Newtons)k is the coefficient of kinetic friction (No unit)N is the normal reaction (in Newtons).
17The direction of the frictional force must be correctly determined by the direction of motion.Sometimes this may not be given and is not obvious.We would have to make an assumption and carryout the analysis. After obtaining the results, we then check that the assumed frictional force is opposite to the motion. If not, then our assumed direction of the frictional force is incorrect and we have to do another analysis.
18Example 7.5A block A of mass 15 kg rests on a rough inclined plane with coefficient = It is connected by an inextensible cord over a smooth light pulley to a hanging mass B of 8 kg. Determine the acceleration of the masses when they are released from rest.10ºB8 kgA15 kg
19The direction of motion for the blocks cannot be obtained by inspection and has to be found as follows.First assume the whole system is in static equilibrium and neglect the effects of friction.10 0B8gTNr15gABlock BFy = T – 8g = 0T = 8g
20Block A Fx = T – 15 x g x sin10º = 8g – 15 x g x sin10º = 52.93 N 0 10 0B8gTNr15gABlock AFx = T – 15 x g x sin10º= 8g – 15 x g x sin10º= N 0Since the tension is larger than the downward component of the weight, block A must move up the plane.
218g – 8a = T --------------- (1) 10 0B8gTNr15gAFaaNow solve the problem as stated, choosing the directions along the motion and draw friction F in.For Block B:F = 8g - T = 8a8g – 8a = T (1)For Block A: Fy = 0Nr – 15 x g x cos10º = 0Nr = NSince there is sliding,F = N= 0.18 x = N
228g - 8a = T ------ (1) from above} 10 0B8gTNr15gAFaastopped hereFx = T – F - 15 x g x sin10º = 15aT – – 15 x g x sin10º = 15aT – = 15a (2)8g - 8a = T (1) from above}Substitute in (2) (8g – 8a) –51.64 = 15a8g – = 23aa = m/s2Note: The positive sign obtain for the acceleration confirms that the direction of motion is correct.
23Example 7.6As shown in the figure, block A has a mass of 30 kg and block B has a mass of 15 kg. If the magnitude of the force P is 250 N, determine the acceleration of block A and the tension in the cord. = 0.3 = 0.430ºPBA
24a a For Block A, Fy = 0 N1 – 30 x g x cos30º = 0 N1 = 254.9 N PN1F1T30º30gN2F215gBlock BaaFor Block A, Fy = 0N1 – 30 x g x cos30º = 0N1 = NF1 = N1= 0.4 x = 102 NFx = maP – F x g x sin30º - T = 30aT = a (1)
25a a For Block B, Fy = 0 N2 - N1 - 15 x g x cos30º = 0 N2 = 382.3 N Block APN1F1T30º30gN2F215gBlock BaaFor Block B, Fy = 0N2 - N x g x cos30º = 0N2 = NF2 = N2= 0.3 x = NFx = maT - F1 - F x g x sin30º = 15aT = a (2)
26Same tension T as the pulley is smooth, 295.15 - 30a = 290.28 + 15a Block APN1F1T30º30gN2F215gBlock BaaFrom above,T = a (1)T = a (2)Same tension T as the pulley is smooth,a = aa = 0.11 m/s2Sub into (1):T = x 0.11= N
277.6 Centripetal and centrifugal forces When a body moves in a circle, there is a centripetalacceleration towards the center of rotation. Newton’s2nd law states that there is a centripetal force,directed towards the center of rotation, causing thisacceleration.F = manFrom Chapter 6 (section 6.2.6):an = vor an = v2/r (Since = v/r)or an = r (Since v = r)( Note:- Total acceleration aT2 = an2 + at2 )
28For a particle tied to a string and swung round in a circle, the tension in the string provides thecentripetal force. If the string breaks, the particlecannot continue in the circular motion.Centrifugal force is a fictitious force but are useful for engineers. If we imagine we are the particlebeing swung round, then there seem to be a force,(a centrifugal force), throwing us away from thecenter of the circular path.
29Example 7.7A string of length 1 m has a maximum tensile strength of 600 N. A 1.2 kg mass is tied to one end. The other end is held and the mass swung round in a horizontal circle on a smooth table. Determinei) the maximum angular velocity possible without breaking the string,ii) the velocity at which the mass flies off when the string breaks.i) Let the tension in the string be T.F = manT = mr2 (an = v = r2)600 = 1.2 x 1 x 2 = rad/sii) The tangential velocity,v = r= 1 x = m/s
30the tension in the string, the linear speed of the plane, Example 7.8A toy aeroplane of mass 0.3 kg is suspended from a 1.2 m string and flies in a circular path of diameter 900 mm. Determinethe tension in the string,the linear speed of the plane,the rpm of the motion.(rpm - Revolution Per Minute)1.2 m0.3gT0.45 mcos = 0.45 / 1.2 =
31In the vertical direction, Fy = 0 (no motion) T sin – 0.3g = 0 cos = 0.45 / 1.2 = Solution:In the vertical direction,Fy = (no motion)T sin – 0.3g = 0T = NIn the horizontal direction,Fx = T cos = x = NThis force causes the plane to have a centripetal acceleration towards the center.
32End of Chapter 7 cos = 0.45 / 1.2 = 67.98 F = man 0.3gT0.45 mcos = 0.45 / 1.2 = F = manF = mv2/r (an = v = v2/r)1.191 = 0.3 x v2 / 0.45v = m/sWe have, v = r1.337 = 0.45 x = rad/srpm = x 60 / 2=End of Chapter 7