{"id":310,"date":"2018-01-26T09:32:00","date_gmt":"2018-01-26T01:32:00","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=310"},"modified":"2018-01-26T09:32:00","modified_gmt":"2018-01-26T01:32:00","slug":"two-bodies-with-the-same-magnitude-of-acceleration-application-of-newtons-law-of-motion-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/two-bodies-with-the-same-magnitude-of-acceleration-application-of-newtons-law-of-motion-problems-and-solutions.htm","title":{"rendered":"Two bodies with the same magnitude of acceleration \u2013 Application of Newton’s law of motion problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. Two masses m<\/span><\/span>1 <\/span><\/sub>= 2 kg and m<\/span><\/span>2<\/span><\/sub> = 5 kg are on inclined plane and are connected together by a string as shown in the figure. The coefficient of the kinetic friction between m<\/span><\/span>1<\/span><\/span><\/sub> and incline is 0.2 and the coefficient of the kinetic friction<\/a> between m<\/span><\/span>2<\/span><\/sub> and incline is 0.1. <\/span><\/span><\/p>\n

(a) Determine their acceleration<\/a> <\/span><\/span><\/p>\n

(b) Determine the tension force<\/span><\/span><\/p>\n

\"Two<\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Mass<\/a> 1 (m<\/span><\/span>1<\/span><\/sub>)<\/span><\/span> = 2 kg<\/span><\/span><\/p>\n

Mass 2 (m<\/span><\/span>2<\/span><\/sub>)<\/span><\/span> = 4 kg<\/span><\/span><\/p>\n

Coefficient of the kinetic friction between m<\/span><\/span>1<\/span><\/span><\/sub> and inclined plane<\/a> (\u03bc<\/span><\/span>k1<\/span><\/sub>)<\/span><\/span> = 0.2<\/span><\/span><\/p>\n

Coefficient of the kinetic friction between m<\/span><\/span>2<\/span><\/span><\/sub> and inclined plane (\u03bc<\/span><\/span>k2<\/span><\/sub>)<\/span><\/span> = 0.1<\/span><\/span><\/p>\n

Acceleration due to gravity<\/a> (g) = 9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

a) The magnitude and direction of the acceleration<\/span><\/span><\/p>\n

\"Two<\/p>\n

w<\/span><\/span>1<\/span><\/sub> = weight<\/a> 1 = m<\/span><\/span>1 <\/span><\/span><\/sub>g = (2 kg)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 19.6 Newton<\/span><\/span><\/p>\n

w<\/span><\/span>1x<\/span><\/sub> = w<\/span><\/span>1<\/span><\/sub> sin 30o<\/sup> = (19.6 N)(0.5) = 9.8 Newton<\/span><\/span><\/p>\n

w<\/span><\/span>1y<\/span><\/sub> = w<\/span><\/span>1<\/span><\/sub> cos 30<\/span><\/span>o <\/span><\/span><\/sup>= (19.6 N)(0.87) = 17 Newton<\/span><\/span><\/p>\n

N<\/span><\/span>1 <\/span><\/sub>= The normal force<\/a> on m1<\/sub> = w<\/span><\/span>1y<\/span><\/sub> = 17 Newton<\/span><\/span><\/p>\n

F<\/span><\/span>k1<\/span><\/sub> = The force of the kinetic friction on m<\/span><\/span>1<\/span><\/span><\/sub> = \u03bc<\/span><\/span>k1 <\/span><\/sub>N<\/span><\/span>1<\/span><\/sub> = (0.2)(17 N) = 3.4 Newton<\/span><\/span><\/p>\n

———<\/span><\/span><\/p>\n

w<\/span><\/span>2<\/span><\/sub> = weight 2 = m<\/span><\/span>2 <\/span><\/sub>g = (4 kg)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 39.2 Newton<\/span><\/span><\/p>\n

w<\/span><\/span>2x<\/span><\/sub> = w<\/span><\/span>2 <\/span><\/sub>sin 60<\/span><\/span>o<\/span><\/span><\/sup> = (39.2 N)(0.87) = 34.1 Newton<\/span><\/span><\/p>\n

w<\/span><\/span>2y<\/span><\/sub> = w<\/span><\/span>2<\/sub> <\/span><\/span>cos 60<\/span><\/span>o <\/span><\/span><\/sup>= (39.2 N)(0.5) = 19.6 Newton<\/span><\/span><\/p>\n

N<\/span><\/span>2<\/span><\/sub> = The normal force on m<\/span><\/span>2<\/span><\/sub> = w<\/span><\/span>2y <\/span><\/sub>= 19.6 Newton<\/span><\/span><\/p>\n

F<\/span><\/span>k2 <\/span><\/sub>= The force of the kinetic friction on m<\/span><\/span>2<\/span><\/sub> = \u03bc<\/span><\/span>k2<\/span><\/sub> N<\/span><\/span>2<\/span><\/sub> = (0.1)(19.6 N) = 1.96 Newton<\/span><\/span><\/p>\n

———<\/span><\/span><\/p>\n

The magnitude of the acceleration :<\/span><\/span><\/p>\n

\u2211F<\/span><\/span>x<\/span><\/sub> = m a<\/span><\/span>x<\/span><\/sub><\/p>\n

w<\/span><\/span>2x <\/span><\/sub>> w<\/span><\/span>1x<\/span><\/sub> so direction of the acceleration is the same as direction of w<\/span><\/span>2x<\/span><\/sub>. <\/span><\/span><\/p>\n

Forces which points along acceleration is positive and forces which has opposite direction with acceleration is negative.<\/span><\/span><\/p>\n

w<\/span><\/span>2x <\/span><\/sub>\u2013 F<\/span><\/span>k2<\/span><\/sub> \u2013 T<\/span><\/span>2<\/span><\/sub> + T<\/span><\/span>1<\/span><\/sub> \u2013 w<\/span><\/span>1x<\/span><\/sub> \u2013 F<\/span><\/span>k1 <\/span><\/sub>= (m<\/span><\/span>1<\/span><\/sub> + m<\/span><\/span>2<\/span><\/sub>) a<\/span><\/span>x<\/span><\/sub><\/p>\n

w<\/span><\/span>2x<\/span><\/sub> \u2013 F<\/span><\/span>k2<\/span><\/sub> \u2013 w<\/span><\/span>1x<\/span><\/sub> \u2013 F<\/span><\/span>k1<\/span><\/sub> = (m<\/span><\/span>1<\/span><\/sub> + m<\/span><\/span>2<\/span><\/sub> ) a<\/span><\/span>x<\/span><\/sub><\/p>\n

34.1 N – 1.96 N – 9.8 N – 3.4 N = (2 kg + 4 kg) a<\/span><\/span>x<\/span><\/span><\/sub><\/p>\n

18.94 N = (6 kg) a<\/span><\/span>x<\/span><\/span><\/sub><\/p>\n

a<\/span><\/span>x<\/span><\/span><\/sub> = 18.94 N : 6 kg<\/span><\/span><\/p>\n

a<\/span><\/span>x <\/span><\/span><\/sub>= 3.16 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

Magnitude of the acceleration = 3.16 m\/s<\/span><\/span>2<\/span><\/span><\/sup> . Direction of the acceleration = direction of T<\/span><\/span>1 <\/span><\/sub>= direction of w<\/span><\/span>2x<\/span><\/sub><\/p>\n

b) Magnitude of the tension force<\/span><\/span><\/p>\n

Apply Newton’s second law on the object 2 :<\/span><\/span><\/p>\n

w<\/span><\/span>2x <\/span><\/sub>\u2013 F<\/span><\/span>k2 <\/span><\/sub>\u2013 T<\/span><\/span>2<\/span><\/sub> = m<\/span><\/span>2 <\/span><\/sub>a<\/span><\/span>x<\/span><\/sub><\/p>\n

34.1 N \u2013 1.96 N \u2013 T<\/span><\/span>2 <\/span><\/sub>= (4 kg)(3.16 m\/s<\/span><\/span>2<\/span><\/span><\/sup>)<\/span><\/span><\/p>\n

32.14 N – T<\/span><\/span>2<\/span><\/sub> = 12.64 N<\/span><\/span><\/p>\n

T<\/span><\/span>2 <\/span><\/sub>= 32.14 N – 12.64 N = 19.5 Newton<\/span><\/span><\/p>\n

The tension force = T = T<\/span><\/span>1<\/span><\/sub> = T<\/span><\/span>2<\/span><\/sub> = 19.5 Newton<\/span><\/span><\/p>\n

<\/p>\n

2. m<\/span><\/span>1 <\/span><\/sub>= 4 kg, m<\/span><\/span>2<\/span><\/sub> = 2 kg. Determine (a) magnitude and direction of the acceleration (b) Magnitude of the tension force which connecting m<\/span><\/span>1<\/span><\/sub> and m<\/span><\/span>2<\/span><\/sub> (c) magnitude of the tension force which connecting pulley and roof.<\/span><\/span><\/p>\n

\"Two<\/p>\n

Solution<\/span><\/span><\/p>\n

\"Two<\/p>\n

w<\/span><\/span>1<\/span><\/sub> = m<\/span><\/span>1 <\/span><\/sub>g = (4 kg)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 39.2 Newton<\/span><\/span><\/p>\n

w<\/span><\/span>2 <\/span><\/sub>= m<\/span><\/span>2<\/span><\/sub> g = (2 kg)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 19.6 Newton<\/span><\/span><\/p>\n

a) Magnitude and direction of the acceleration<\/span><\/span><\/p>\n

\u2211F<\/span><\/span>y<\/span><\/sub> = m a<\/span><\/span>y<\/span><\/sub><\/p>\n

w<\/i><\/span><\/span>1 <\/i><\/span><\/sub>> w<\/i><\/span><\/span>2 <\/i><\/span><\/sub>so the direction of the object is same as the direction of the weight 1 (<\/i><\/span><\/span>w<\/i><\/span><\/span>1<\/i><\/span><\/sub>)<\/i><\/span><\/span>. Forces which has the same direction with acceleration is positive and forces which has opposite direction with acceleration is negative.<\/i><\/span><\/span><\/p>\n

w<\/span><\/span>1 <\/span><\/sub>\u2013 T<\/span><\/span>1<\/span><\/sub> + T<\/span><\/span>2<\/span><\/sub> \u2013 w<\/span><\/span>2 <\/span><\/sub>= (m<\/span><\/span>1<\/span><\/sub> + m<\/span><\/span>2<\/span><\/sub>) a<\/span><\/span>y<\/span><\/sub><\/p>\n

w<\/span><\/span>1<\/span><\/sub> \u2013 w<\/span><\/span>2<\/span><\/sub> = (m<\/span><\/span>1<\/span><\/sub> + m<\/span><\/span>2<\/span><\/sub>) a<\/span><\/span>y<\/span><\/span><\/sub><\/p>\n

39.2 N – 19.6 N = (4 kg + 2 kg) a<\/span><\/span>y<\/span><\/sub><\/p>\n

19.6 N = (6 kg) a<\/span><\/span>y<\/span><\/sub><\/p>\n

a<\/span><\/span>y <\/span><\/sub>= 19.6 N : 6 kg<\/span><\/span><\/p>\n

a<\/span><\/span>y<\/span><\/sub> = 3.26 m\/s2<\/sup><\/span><\/span><\/p>\n

Magnitude of acceleration = 3.26 m\/s<\/span><\/span>2<\/span><\/span><\/sup>. Direction of acceleration = direction of w<\/span><\/span>1<\/span><\/span><\/sub> .<\/span><\/span><\/p>\n

b) Magnitude of tension force which connecting m<\/span><\/span>1<\/span><\/sub> and m<\/span><\/span>2<\/span><\/sub><\/p>\n

Apply Newton’s second law<\/a> on m<\/span><\/span>2<\/span><\/span><\/sub> :<\/span><\/span><\/p>\n

\u2211Fy<\/sub> = m a<\/span><\/span>y<\/span><\/sub><\/p>\n

w<\/span><\/span>1<\/span><\/sub> \u2013 T<\/span><\/span>1<\/span><\/sub> = m<\/span><\/span>1 <\/span><\/sub>a<\/span><\/span>y<\/span><\/sub><\/p>\n

39.2 N – T<\/span><\/span>1 <\/span><\/sub>= (4 kg)( 3.26 m\/s<\/span><\/span>2<\/span><\/span><\/sup>)<\/span><\/span><\/p>\n

39.2 N \u2013 T<\/span><\/span>1<\/span><\/sub> = 13.04 N<\/span><\/span><\/p>\n

T<\/span><\/span>1 <\/span><\/sub>= 39.2 N – 13.04 N<\/span><\/span><\/p>\n

T<\/span><\/span>1<\/span><\/sub> = 26.16 Newton<\/span><\/span><\/p>\n

Magnitude of the tension force which connection objects = T = T<\/span><\/span>1<\/span><\/sub> = T<\/span><\/span>2<\/span><\/sub> = 26.16 Newton<\/span><\/span><\/p>\n

c) Magnitude of the tension force which connecting pulley and roof.<\/span><\/span><\/p>\n

\"TwoPulley is at rest :<\/span><\/span><\/p>\n

\u2211F<\/span><\/span>y <\/span><\/span><\/sub>= m a<\/span><\/span>y<\/span><\/sub> —— a<\/span><\/span>y <\/span><\/sub>= 0<\/span><\/span><\/p>\n

\u2211F<\/span><\/span>y <\/span><\/sub>= 0<\/span><\/span><\/p>\n

Upward force are positive, downward forces are negative :<\/span><\/span><\/p>\n

T<\/span><\/span>3<\/span><\/sub> \u2013 T<\/span><\/span>1<\/span><\/sub> \u2013 T<\/span><\/span>2<\/span><\/sub> = 0<\/span><\/span><\/p>\n

T<\/span><\/span>3 <\/span><\/sub>= T<\/span><\/span>1<\/span><\/sub> + T<\/span><\/span>2<\/span><\/sub><\/p>\n

T<\/span><\/span>1<\/span><\/sub> and T<\/span><\/span>2<\/span><\/sub> have the same magnitude<\/span><\/span>, T<\/span><\/span>1<\/span><\/sub> = T<\/span><\/span>2<\/span><\/sub> = T = 26.16 N :<\/span><\/span><\/p>\n

T<\/span><\/span>3<\/span><\/sub> = 2T = 2(26.16 N) = 52.32 Newton<\/span><\/span><\/p>\n

<\/p>\n

3. Block 1 (m<\/span><\/span>1<\/span><\/sub> = 10 kg) and block 2 (m<\/span><\/span>2 <\/span><\/sub>= 15 kg) connected by a cord over frictionless pulley. Coefficient of the static friction between the block 2 with incline = 0.6. The coefficient of the kinetic friction between the block 2 with incline = 0.42. Determine (a) The magnitude of the minimum force F exerted on the objects so the objects accelerated upward (b) Determine the magnitude of the tension force.<\/span><\/span><\/p>\n

\"Two<\/p>\n

Solution<\/span><\/span><\/p>\n

\"Two<\/p>\n

w<\/span><\/span>1<\/span><\/sub> = The weight of the block 1 = m<\/span><\/span>1<\/span><\/sub> g = (10 kg)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 98 Newton<\/span><\/span><\/p>\n

w<\/span><\/span>2<\/span><\/sub> = The weight of the block 2 = m<\/span><\/span>2<\/span><\/sub> g = (15 kg)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 147 Newton<\/span><\/span><\/p>\n

w<\/span><\/span>2y<\/span><\/sub> = w<\/span><\/span>2<\/span><\/sub> cos 30<\/span><\/span>o<\/span><\/span><\/sup> = (147 N)(0.87) = 127.89 Newton<\/span><\/span><\/p>\n

w<\/span><\/span>2x<\/span><\/sub> = w<\/span><\/span>2 <\/span><\/sub>sin 30<\/span><\/span>o<\/span><\/span><\/sup> = (147 N)(0.5) = 73.5 Newton<\/span><\/span><\/p>\n

N<\/span><\/span>2<\/span><\/sub> = The normal force on the block 2 = w<\/span><\/span>2y <\/span><\/sub>= 127.89 Newton<\/span><\/span><\/p>\n

F<\/span><\/span>k2<\/span><\/sub> = The force of the kinetic friction on the block 2 = \u03bc<\/span><\/span>k2<\/span><\/sub> N<\/span><\/span>2<\/span><\/sub> = (0.42)(127.89 N) = 53.7 Newton<\/span><\/span><\/p>\n

F<\/span><\/span>s2<\/span><\/sub> = The force of the static friction on the block 2 = \u03bc<\/span><\/span>s2<\/span><\/sub> N<\/span><\/span>2<\/span><\/sub> = (0.6)(127.89 N) = 76.7 Newton<\/span><\/span><\/p>\n

a) The magnitude of the minimum force F exerted on the objects so the objects accelerated upward <\/b><\/span><\/span><\/p>\n

\u2211F<\/span><\/span>x<\/span><\/sub> = m a<\/span><\/span>x<\/span><\/sub> —— a<\/span><\/span>x<\/span><\/span><\/sub> = 0<\/span><\/span><\/p>\n

\u2211F<\/span><\/span>x<\/span><\/sub> = 0<\/span><\/span><\/p>\n

Upward forces and rightward forces are positive, downward forces and leftward forces are negative.<\/span><\/span><\/p>\n

F \u2013 F<\/span><\/span>k2<\/span><\/sub> \u2013 w<\/span><\/span>2x<\/span><\/sub> \u2013 w<\/span><\/span>1<\/span><\/sub> – T<\/span><\/span>2<\/span><\/sub> + T<\/span><\/span>1<\/span><\/sub> = 0<\/span><\/span><\/p>\n

F \u2013 F<\/span><\/span>k2<\/span><\/sub> \u2013 w<\/span><\/span>2x <\/span><\/sub>\u2013 w<\/span><\/span>1<\/span><\/sub> = 0<\/span><\/span><\/p>\n

F = F<\/span><\/span>k2 <\/span><\/sub>+ w<\/span><\/span>2x<\/span><\/sub> + w<\/span><\/span>1<\/span><\/sub><\/p>\n

F = 53.7 N + 73.5 N + 98 N<\/span><\/span><\/p>\n

F = 225.2 Newton<\/span><\/span><\/p>\n

b) The magnitude of the tension force<\/b><\/span><\/span><\/p>\n

Apply Newton’s law of the motion on the block 1 :<\/span><\/span><\/p>\n

\u2211F<\/span><\/span>y<\/span><\/sub> = m a<\/span><\/span>y <\/span><\/sub>—— a<\/span><\/span>y <\/span><\/sub>= 0<\/span><\/span><\/p>\n

\u2211F<\/span><\/span>y<\/span><\/sub> = 0<\/span><\/span><\/p>\n

T<\/span><\/span>1<\/span><\/sub> \u2013 w<\/span><\/span>1<\/span><\/sub> = 0<\/span><\/span><\/p>\n

T<\/span><\/span>1<\/span><\/sub> = w<\/span><\/span>1<\/span><\/sub> = 98 Newton<\/span><\/span><\/p>\n

Apply Newton’s law of the motion on the block 2 :<\/span><\/span><\/p>\n

F – F<\/span><\/span>k2<\/span><\/sub> \u2013 w<\/span><\/span>2x<\/span><\/sub> \u2013 T<\/span><\/span>2<\/span><\/sub> = 0<\/span><\/span><\/p>\n

T<\/span><\/span>2 <\/span><\/sub>= F – F<\/span><\/span>k2 <\/span><\/sub>\u2013 w<\/span><\/span>2x<\/span><\/sub><\/p>\n

T<\/span><\/span>2 <\/span><\/sub>= 225.2 N – 53.7 N \u2013 73.5 N<\/span><\/span><\/p>\n

T<\/span><\/span>2<\/span><\/sub> = 98 Newton<\/span><\/span><\/p>\n

Magnitude of the tension force = T<\/span><\/span>1<\/span><\/sub> = T<\/span><\/span>2<\/span><\/sub> = T = 98 Newton<\/span><\/span><\/p>\n

<\/p>\n

4. Block 1 (m<\/span><\/span>1<\/span><\/sub> = 16 kg) lies on a horizontal surface and the block 2 (m<\/span><\/span>2<\/span><\/sub> = 12 kg) lies on a smooth inclined plane, connected by a cord that passes over a small, frictionless pulley. Block 3 (m<\/span><\/span>3<\/span><\/sub> = 5 kg) lies on the block 2. The coefficient of the kinetic friction between the block 2 and the horizontal surface is 0,4. The coe<\/span><\/span>f<\/span><\/span>ficient of the static friction between the block 2 with the block 3 is 0,3. <\/span><\/span><\/p>\n

(a) <\/span><\/span>When the system is released from rest, the block 3 and the block 2 still slide together ?<\/span><\/span><\/p>\n

(b) <\/span><\/span>If there is no block 3, what is the acceleration of the block 1 and the block 2 ? <\/span><\/span><\/p>\n

\"Two<\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

a) <\/span><\/span>When the system is released from rest, the block 3 and the block 2 still slide together?<\/span><\/span><\/p>\n

\"Two<\/p>\n

w<\/span><\/span>1<\/span><\/sub> = The <\/span><\/span>weight of the block <\/span><\/span>1 = m<\/span><\/span>1<\/span><\/sub> g = (16 kg)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 156.8 Newton<\/span><\/span><\/p>\n

w<\/span><\/span>1x<\/span><\/sub> = w<\/span><\/span>1<\/span><\/span><\/sub> sin 60<\/span><\/span>o<\/span><\/span><\/sup> = (156.8 N)(0.87) = 136.4 Newton<\/span><\/span><\/p>\n

w<\/span><\/span>1y<\/span><\/sub> = w<\/span><\/span>1<\/span><\/span><\/sub> cos 60<\/span><\/span>o<\/span><\/span><\/sup> = (156.8 N)(0.5) = 78.4 Newton<\/span><\/span><\/p>\n

N<\/span><\/span>1<\/span><\/sub> = The <\/span><\/span>normal force exerted on the block 1 by the inclined plane <\/span><\/span>= w<\/span><\/span>1y<\/sub> <\/span><\/span>= 78.4 Newton<\/span><\/span><\/p>\n

w<\/span><\/span>3<\/span><\/sub> = The <\/span><\/span>weight of the block <\/span><\/span>3 = m<\/span><\/span>3<\/span><\/sub> g = (5 kg)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 49 Newton<\/span><\/span><\/p>\n

N<\/span><\/span>23<\/span><\/sub> = The <\/span><\/span>normal force exerted on the block 3 bythe\u00a0 block 2 <\/span><\/span>= w<\/span><\/span>3 <\/span><\/sub>= 49 Newton<\/span><\/span><\/p>\n

N<\/span><\/span>32<\/span><\/sub> = The n<\/span><\/span>ormal force exerted on the block 2 by the block 3 <\/span><\/span>= N<\/span><\/span>23<\/span><\/sub> = w<\/span><\/span>3<\/span><\/sub> = 49 Newton<\/span><\/span><\/p>\n

(N<\/span><\/span>23 <\/span><\/sub>and <\/span><\/span>N<\/span><\/span>32<\/span><\/sub> are action-reaction pair<\/span><\/span>)<\/span><\/span><\/p>\n

F<\/span><\/span>s23<\/span><\/sub> = The <\/span><\/span>force of the static friction exerted on the block 3 by the block 2 <\/span><\/span>= \u03bc<\/span><\/span>s<\/span><\/sub> N<\/span><\/span>23 <\/span><\/sub>= (0.3)(49 N) = 14.7 <\/span><\/span>Newton<\/span><\/span><\/p>\n

F<\/span><\/span>s32 <\/span><\/sub>= The <\/span><\/span>force of the static friction exerted on th block 2 by the block 3 <\/span><\/span>= Fs<\/sub><\/span><\/span>23<\/span><\/sub> = 14.7 Newton<\/span><\/span><\/p>\n

(F<\/span><\/span>s23<\/span><\/sub> and <\/span><\/span>F<\/span><\/span>s32<\/span><\/sub> are action-reaction pair<\/span><\/span>)<\/span><\/span><\/p>\n

w<\/span><\/span>2<\/span><\/sub> = The <\/span><\/span>weight of the block 2 <\/span><\/span>= m<\/span><\/span>2<\/span><\/sub> g = (12 kg)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 117.6 Newton<\/span><\/span><\/p>\n

N<\/span><\/span>2<\/span><\/sub> = The <\/span><\/span>normal force exerted on the object 2 by the horizontal surface <\/span><\/span>= w<\/span><\/span>2<\/span><\/sub> + N<\/span><\/span>32<\/span><\/sub> = 117.6 Newton + 49<\/span><\/span><\/p>\n

Newton = 166.6 Newton<\/span><\/span><\/p>\n

F<\/span><\/span>k2 <\/span><\/sub>= The <\/span><\/span>force of the kinetic friction on the block 2 <\/span><\/span>= \u03bc<\/span><\/span>k<\/span><\/sub> N<\/span><\/span>2<\/span><\/sub> = (0.4)(166.6 N) = 66.64 Newton<\/span><\/span><\/p>\n

Apply Newton’s law of motion on the block 3 : <\/span><\/span><\/p>\n

\u2211F<\/span><\/span>x<\/span><\/sub> = m a<\/span><\/span>x<\/span><\/sub><\/p>\n

F<\/span><\/span>s23 <\/span><\/sub>=m<\/span><\/span>3 <\/span><\/sub>a<\/span><\/span>x<\/span><\/sub><\/p>\n

—–> F<\/span><\/span>s23<\/span><\/sub> = \u03bc<\/span><\/span>s <\/span><\/sub>N<\/span><\/span>23 <\/span><\/sub>= \u03bc<\/span><\/span>s <\/span><\/sub>w<\/span><\/span>3 <\/span><\/sub>= \u03bc<\/span><\/span>s <\/span><\/sub>m<\/span><\/span>3 <\/span><\/sub>g<\/span><\/span><\/p>\n

\u03bc<\/span><\/span>s<\/span><\/sub> m<\/span><\/span>3 <\/span><\/sub>g = m<\/span><\/span>3 <\/span><\/sub>a<\/span><\/span>x<\/span><\/sub><\/p>\n

\u03bc<\/span><\/span>s<\/span><\/sub> g = a<\/span><\/span>x<\/span><\/sub><\/p>\n

a<\/span><\/span>x<\/span><\/sub> = (0.3)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 2.94 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

The maximum acceleration of the block 3 so that the block 3 and the block 2 still slide together is 2.94 m\/s<\/span><\/span>2<\/span><\/span><\/sup>. <\/span><\/span><\/p>\n

Now we calculate the magnitude of the system’s acceleration after released from rest. <\/span><\/span><\/p>\n

The direction of the block displacement = the direction of the block’s acceleration = the direction of T<\/span><\/span>2 <\/span><\/sub>= the direction of w<\/span><\/span>1x<\/span><\/sub>.<\/span><\/span><\/p>\n

\u2211F<\/span><\/span>x<\/span><\/sub> = m a<\/span><\/span>x<\/span><\/sub><\/p>\n

w<\/span><\/span>1x<\/span><\/sub> \u2013 T<\/span><\/span>1 <\/span><\/sub>+ T<\/span><\/span>2 <\/span><\/sub>\u2013 F<\/span><\/span>k2<\/span><\/sub> \u2013 F<\/span><\/span>s32<\/span><\/sub> + F<\/span><\/span>s23<\/span><\/sub> = (m<\/span><\/span>1<\/span><\/sub> + m<\/span><\/span>2<\/span><\/sub> + m<\/span><\/span>3<\/span><\/sub>) a<\/span><\/span>x<\/span><\/sub><\/p>\n

w<\/span><\/span>1x<\/span><\/sub> \u2013 F<\/span><\/span>k2<\/span><\/sub> = (m<\/span><\/span>1<\/span><\/sub> + m<\/span><\/span>2 <\/span><\/sub>+ m<\/span><\/span>3<\/span><\/sub> ) a<\/span><\/span>x<\/span><\/sub><\/p>\n

136.4 N \u2013 66.64 N = (16 kg + 12 kg + 5 kg) a<\/span><\/span>x<\/span><\/sub><\/p>\n

69.76 N = (33 kg) a<\/span><\/span>x<\/span><\/sub><\/p>\n

a<\/span><\/span>x<\/span><\/sub> = 2.11 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

ax<\/sub> is positive, means direction of the block displacement or the direction of the acceleration is same as direction of T<\/span><\/span>2<\/span><\/sub> or direction of w<\/span><\/span>1x<\/span><\/sub>. <\/span><\/span><\/p>\n

The magnitude of the acceleration is <\/span><\/span>2.11 m\/s<\/span><\/span>2<\/span><\/span><\/sup> , l<\/span><\/span>ower than <\/span><\/span>2.94 m\/s<\/span><\/span>2 <\/span><\/span><\/sup>so we can conclude that block 3 and block 2 still slide together after released from rest. <\/span><\/span><\/p>\n

b) <\/span><\/span>The magnitude of the acceleration of the block 1 and the block 2<\/span><\/span><\/p>\n

\u2211F<\/span><\/span>x <\/span><\/sub>= m a<\/span><\/span>x<\/span><\/sub><\/p>\n

w<\/span><\/span>1x<\/span><\/sub> \u2013 F<\/span><\/span>k2 <\/span><\/sub>= (m<\/span><\/span>1<\/span><\/sub> + m<\/span><\/span>2<\/span><\/sub>) ax<\/sub><\/span><\/span><\/p>\n

—–> F<\/span><\/span>k2<\/span><\/sub> = \u03bc<\/span><\/span>k<\/span><\/sub> N<\/span><\/span>2 <\/span><\/sub>= \u03bc<\/span><\/span>k<\/span><\/sub> w<\/span><\/span>2<\/sub> <\/span><\/span>= \u03bc<\/span><\/span>k<\/span><\/sub> m<\/span><\/span>2 <\/span><\/sub>g = (0.4)(12 kg)(9.8 m\/s<\/span><\/span>2<\/span><\/span><\/sup>) = 47.04 Newton<\/span><\/span><\/p>\n

136.4 N – 47.04 N = (16 kg + 12 kg) a<\/span><\/span>x<\/span><\/sub><\/p>\n

89.36 N = (28 kg) ax<\/sub><\/span><\/span><\/p>\n

a<\/span><\/span>x <\/span><\/sub>= 89.36 N : 28 kg = 3.19 m\/s<\/span><\/span>2<\/span><\/span><\/sup><\/p>\n

[wpdm_package id=’493′]<\/p>\n

    \n
  1. Mass and weight<\/a><\/li>\n
  2. Normal force<\/a><\/li>\n
  3. Newton’s second law of motion<\/a><\/li>\n
  4. Friction force<\/a><\/li>\n
  5. Motion on the horizontal surface without friction force<\/a><\/li>\n
  6. The motion of two bodies with the same acceleration on the rough horizontal surface with the friction force<\/a><\/li>\n
  7. Motion on the inclined plane without friction force<\/a><\/li>\n
  8. Motion on the rough inclined plane with the friction force<\/a><\/li>\n
  9. Motion in an elevator<\/a><\/li>\n
  10. The motion of bodies connected by cord and pulley<\/a><\/li>\n
  11. Two bodies with the same magnitude of accelerations<\/a><\/li>\n
  12. Rounding a flat curve – dynamics of circular motion<\/a><\/li>\n
  13. Rounding a banked curve – dynamics of circular motion<\/a><\/li>\n
  14. Uniform motion in a horizontal circle<\/a><\/li>\n
  15. Centripetal force in uniform circular motion<\/a><\/li>\n<\/ol>\n

    <\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

    1. Two masses m1 = 2 kg and m2 = 5 kg are on inclined plane and are connected together by a string as shown in the figure. The coefficient of the kinetic friction between m1 and incline is 0.2 and the coefficient of the kinetic friction between m2 and incline is 0.1. (a) Determine … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Two bodies with the same magnitude of acceleration \u2013 Application of Newton's law of motion problems and solutions","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[3],"tags":[],"class_list":["post-310","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/310","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=310"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/310\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=310"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=310"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=310"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}