ENGINEERING MECHANICS CHAPTER 7 School of Engineering ENGINEERING MECHANICS CHAPTER 7 Kinetics of Linear Motion – Newton’s 2nd Law Method Click on VIEW and select SLIDE SHOW to view the presentation.

7.1 Brief Kinetics is the study of motion with the forces associated with it. In the earlier chapter under Kinematics, we have studied motion without considering the forces. For analysis of body with forces, Newton’s three laws of motion are applicable. Newton’s First Law: Every body remains at rest or maintains a constant velocity in a straight line unless an unbalanced force acts upon it. .

Newton’s Second Law: A body acted upon by an unbalanced force experiences an acceleration that has the same direction as the force and a magnitude that is directly proportional to the force. Newton’s Third Law: For every force, there is an equal and opposite reaction.

7.2 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 force F = ma for several forces Where F is the force (N) m is the mass (kg) a is the acceleration (m/s2)

7.3 Free body diagram We 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 convention We choose the direction of the resulting motion as the positive direction (note that this is different from Statics) However, any direction can be selected as positive but this must be strictly adhered to throughout the analysis.

6.1.4 Uniform motion A motion is called uniform motion when the acceleration is constant. The three equations of motion with constant acceleration are: v = vi + at v2 = vi2 + 2as s = vit + ½at2 where 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)

a horizontal force of 10 N, towards the right, Example 7.1 A 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 = ma 10 = 5 x a a = 2 m/s2  Fy = 0, (no motion upwards) Nr – 5g = 0 Nr = 49.1 N  Nr 10 5g ( g = 9.81 m/s2 ) F.B.D

ii) Along the x-direction, Fx = - 10cos60 = - 5 N F = ma Nr 60 5g 10 ii) Along the x-direction, Fx = - 10cos60 = - 5 N F = ma - 5 = 5 x a a = - 1 m/s2  a ii) Fy = 0, (no motion upwards) Nr - 5g – 10sin 60 = 0 Nr = 57.71 N 

iii) Along the x-direction, Fx = 10 - 10 cos60 = 5 N F = ma 5 = 5 x a a = 1 m/s2  Fy = 0, (no motion upwards) Nr - 5g – 10sin 60 = 0 Nr = 57.71 N  Nr 60 5g 10 a

In the x-direction, the unbalanced force on the car is Example 7.2 A 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/s2 1500 N 950g N Nr  Solution: In the x-direction, the unbalanced force on the car is Fx = 1500 –950g sin F = ma 1500 –(950 x 9.81 x sin) = 950 x 0.25 Hence,  = 7.780 (Ans.)

The net force is Fx = T – Fr – (500,000 x g x sin) Example 7.3 A 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. x 80 1 Fr = 65 x 500 N T (500 x 1000x g) Nr  Solution: The net force is Fx = T – Fr – (500,000 x g x sin) Since sin = 1/80, Fx = T – 93,812.5 F = ma T – 93,812.5 = 500,000a ----- (1)

Substitute into equation (1), T – 93,812.5 = 500,000a x 80 1 Fr = 65 x 500 N T (500 x 1000x g) Nr  The acceleration of the train, in the x-direction can be obtained from: v = u + at 15,000 / 3600 = 45000 / 3600 + a x 50 a = -0.167 m/s Substitute into equation (1), T – 93,812.5 = 500,000a T – 93,812.5 = 500,000 x (-0.167) T = 10,312.5 N

Two blocks, A and B of masses 10 kg and 6 kg Example 7.4 Two blocks, A and B of masses 10 kg and 6 kg respectively, are connected by an inextensible string over a light frictionless pulley as shown. The system is released from rest. Determine the acceleration of the mass A if the table is smooth. B 6 kg A 10 kg

Nr T 10g 6g A B a Solution: 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 = T F = ma T = 10a ------ (1)

Substitute (1) into (2), T = 10a ------ (1) Nr T 10g 6g A B a For block B, choose positive y direction downward (along the direction of acceleration). Fy = 6g - T F = ma 6g - T = 6a --------- (2) Substitute (1) into (2), T = 10a ------ (1) 6g – 10a = 6a a = 3.678 m/s2

7.5 Friction When the body moves on a rough surface, there will be friction resisting the motion. The friction force equation is (Refer Chapter 5) F = kN where F is the frictional force (in Newtons) k is the coefficient of kinetic friction (No unit) N is the normal reaction (in Newtons).

The 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 carry out 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.

Example 7.5 A block A of mass 15 kg rests on a rough inclined plane with coefficient  = 0.18. 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. 10º B 8 kg A 15 kg

The 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 0 B 8g T Nr 15g A Block B Fy = T – 8g = 0 T = 8g

Block A Fx = T – 15 x g x sin10º = 8g – 15 x g x sin10º = 52.93 N  0 10 0 B 8g T Nr 15g A Block A Fx = T – 15 x g x sin10º = 8g – 15 x g x sin10º = 52.93 N  0 Since the tension is larger than the downward component of the weight, block A must move up the plane.

8g – 8a = T --------------- (1) 10 0 B 8g T Nr 15g A F a a Now solve the problem as stated, choosing the directions along the motion and draw friction F in. For Block B: F = 8g - T = 8a 8g – 8a = T --------------- (1) For Block A: Fy = 0 Nr – 15 x g x cos10º = 0 Nr = 144.91 N Since there is sliding, F = N = 0.18 x 144.91 = 26.08 N

8g - 8a = T ------ (1) from above} 10 0 B 8g T Nr 15g A F a a stopped here Fx = T – F - 15 x g x sin10º = 15a T – 26.08 – 15 x g x sin10º = 15a T – 51.64 = 15a --------- (2) 8g - 8a = T ------ (1) from above} Substitute in (2) (8g – 8a) –51.64 = 15a 8g – 51.64 = 23a a = 1.167 m/s2 Note: The positive sign obtain for the acceleration confirms that the direction of motion is correct.

Example 7.6 As 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  = 0.4 30º P B A

a a For Block A, Fy = 0 N1 – 30 x g x cos30º = 0 N1 = 254.9 N P N1 F1 T 30º 30g N2 F2 15g Block B a a For Block A, Fy = 0 N1 – 30 x g x cos30º = 0 N1 = 254.9 N F1 = N1 = 0.4 x 254.9 = 102 N Fx = ma P – F1 + 30 x g x sin30º - T = 30a T = 295.15 - 30a ------- (1)

a a For Block B, Fy = 0 N2 - N1 - 15 x g x cos30º = 0 N2 = 382.3 N Block A P N1 F1 T 30º 30g N2 F2 15g Block B a a For Block B, Fy = 0 N2 - N1 - 15 x g x cos30º = 0 N2 = 382.3 N F2 = N2 = 0.3 x 382.3 = 114.7 N Fx = ma T - F1 - F2 - 15 x g x sin30º = 15a T = 290.28 + 15a ---------- (2)

Same tension T as the pulley is smooth, 295.15 - 30a = 290.28 + 15a Block A P N1 F1 T 30º 30g N2 F2 15g Block B a a From above, T = 295.15 - 30a ------- (1) T = 290.28 + 15a ---------- (2) Same tension T as the pulley is smooth, 295.15 - 30a = 290.28 + 15a a = 0.11 m/s2 Sub into (1): T = 295.15 - 30 x 0.11 = 291.9 N

7.6 Centripetal and centrifugal forces When a body moves in a circle, there is a centripetal acceleration towards the center of rotation. Newton’s 2nd law states that there is a centripetal force, directed towards the center of rotation, causing this acceleration. F = man From Chapter 6 (section 6.2.6): an = v or an = v2/r (Since  = v/r) or an = r2 (Since v = r) ( Note:- Total acceleration aT2 = an2 + at2 )

For a particle tied to a string and swung round in a circle, the tension in the string provides the centripetal force. If the string breaks, the particle cannot continue in the circular motion. Centrifugal force is a fictitious force but are useful for engineers. If we imagine we are the particle being swung round, then there seem to be a force, (a centrifugal force), throwing us away from the center of the circular path.

Example 7.7 A 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. Determine i) 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 = man T = mr2 (an = v  = r2) 600 = 1.2 x 1 x 2  = 22.36 rad/s ii) The tangential velocity, v = r = 1 x 22.36 = 22.36 m/s

the tension in the string, the linear speed of the plane, Example 7.8 A 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. Determine the tension in the string, the linear speed of the plane, the rpm of the motion.(rpm - Revolution Per Minute) 1.2 m 0.3g T 0.45 m  cos  = 0.45 / 1.2  = 67.98 

In the vertical direction, Fy = 0 (no motion) T sin  – 0.3g = 0 cos  = 0.45 / 1.2  = 67.98  Solution: In the vertical direction, Fy = 0 (no motion) T sin  – 0.3g = 0 T = 3.175 N In the horizontal direction, Fx = T cos  = 3.175 x 0.375 = 1.191 N This force causes the plane to have a centripetal acceleration towards the center.

End of Chapter 7 cos  = 0.45 / 1.2  = 67.98  F = man 0.3g T 0.45 m  cos  = 0.45 / 1.2  = 67.98  F = man F = mv2/r (an = v  = v2/r) 1.191 = 0.3 x v2 / 0.45 v = 1.337 m/s We have, v = r 1.337 = 0.45 x   = 2.971 rad/s rpm = 2.971 x 60 / 2 = 28.37. End of Chapter 7