{"id":2270,"date":"2018-04-29T13:30:56","date_gmt":"2018-04-29T05:30:56","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2270"},"modified":"2023-08-08T13:40:39","modified_gmt":"2023-08-08T13:40:39","slug":"dynamics-of-particles-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/dynamics-of-particles-problems-and-solutions.htm","title":{"rendered":"Dynamics of particles \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Dynamics of particles \u2013 problems and solutions<\/span><\/p>\n

1. Object A with a mass of 6-kg and object B with a mass<\/a> of 4-kg connected by a cord and pulled by a force of F = 60 N, as shown in the figure below. The coefficient of kinetic friction<\/a> between the floor and both objects is 0.5 (tan \u03b8 = \u00be). Acceleration due to gravity<\/a> is 10 m\/s2<\/sup>. What is the magnitude of the tension force?<\/span><\/p>\n

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

Mass of object A (mA<\/sub>) = 6 kg\"Dynamics<\/span><\/p>\n

Mass of object B (mB<\/sub>) = 4 kg<\/span><\/p>\n

Force (F) = 60 Newton<\/span><\/p>\n

The coefficient of kinetic friction between object and floor (\u03bck<\/sub>) = 0.5<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s2\"Dynamics<\/sup><\/span><\/p>\n

Tan \u03b8 = 3\/4<\/span><\/p>\n

Wanted :<\/u> Tension force (T)<\/span><\/p>\n

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

Horizontal component of force F :<\/span><\/p>\n

Fx<\/sub> = F cos \u03b8 <\/span><\/p>\n

Fx<\/sub> = (60)(4\/5) = (4)(12) = 48 N <\/span><\/p>\n

Vertical component of force F :<\/span><\/p>\n

Fy<\/sub> = F sin \u03b8 <\/span><\/p>\n

Fy<\/sub> = (60)(3\/5) = (3)(12) = 36 N <\/span><\/p>\n

The normal force<\/a> on object A :<\/span><\/p>\n

NA<\/sub> = wA <\/sub>= mA<\/sub> g = (6)(10) = 60 N<\/span><\/p>\n

The normal force on object B :<\/span><\/p>\n

NB<\/sub> + Fy<\/sub> \u2013 wB<\/sub> = 0<\/span><\/p>\n

NB<\/sub> + Fy <\/sub>= wB<\/sub><\/span><\/p>\n

NB<\/sub> = wB<\/sub> \u2013 Fy<\/sub> = mB<\/sub> g \u2013 Fy <\/sub>= (4)(10) \u2013 36 = 40 \u2013 36 = 4 N <\/span><\/p>\n

The force of kinetic friction between object A and floor :<\/span><\/p>\n

fkA<\/sub> = \u03bck <\/sub>NA<\/sub> = (0.5)(60) = 30 N<\/span><\/p>\n

The force of kinetic friction between object B and floor :<\/span><\/p>\n

fkB<\/sub> = \u03bck <\/sub>NB<\/sub> = (0.5)(4) = 2 N<\/span><\/p>\n

Calculate the acceleration of both objects :<\/b><\/span><\/p>\n

\u03a3F = m a<\/span><\/p>\n

Fx<\/sub> \u2013 T + T – fkB<\/sub> \u2013 fkA<\/sub> = (mA<\/sub> + mB<\/sub>) a<\/span><\/p>\n

Fx<\/sub> – fkB<\/sub> \u2013 fkA<\/sub> = (mA<\/sub> + mB<\/sub>) a <\/span><\/p>\n

48 \u2013 2 \u2013 30 = (6 + 4) a<\/span><\/p>\n

16 = 10 a<\/span><\/p>\n

a = 16\/10<\/span><\/p>\n

a = 1.6 m\/s2<\/sup><\/span><\/p>\n

Calculate the tension force :<\/b><\/span><\/p>\n

Object A :<\/span><\/p>\n

\u03a3F = m a<\/span><\/p>\n

TA<\/sub> – fkA <\/sub>= mA<\/sub> a<\/span><\/p>\n

TA<\/sub> – 30 = (6)(1.6)<\/span><\/p>\n

TA<\/sub> \u2013 30 = 9.6<\/span><\/p>\n

TA <\/sub>= 9.6 + 30 = 39.6 N<\/span><\/p>\n

Object B :<\/span><\/p>\n

\u03a3F = m a<\/span><\/p>\n

Fx <\/sub>\u2013 fkB <\/sub>\u2013 TB <\/sub>= mB<\/sub> a<\/span><\/p>\n

48 – 2 – TB <\/sub>= (4)(1.6)<\/span><\/p>\n

46 – TB <\/sub>= 6.4<\/span><\/p>\n

46 \u2013 6.4 = TB <\/sub><\/span><\/p>\n

TB <\/sub>= 39.6 N<\/span><\/p>\n

2. If the coefficient of kinetic friction between both blocks and floor is 0.2 , what is the acceleration<\/a> of both objects ? (cos 37o<\/sup> = 0,8, sin 37o <\/sup>= 0,6)<\/span><\/p>\n

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

Mass of object A (mA<\/sub>) = 4 kg\"Dynamics<\/span><\/p>\n

Mass of object B (mB<\/sub>) = 2 kg<\/span><\/p>\n

Force (F) = 30 Newton<\/span><\/p>\n

The coefficient of kinetic friction between object and floor (\u03bck<\/sub>) = 0.2<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s2\"Dynamics<\/sup><\/span><\/p>\n

cos 37o<\/sup> = 0.8<\/span><\/p>\n

sin 37o <\/sup>= 0.6<\/span><\/p>\n

Wanted :<\/u> Acceleration of both objects<\/span><\/p>\n

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

The horizontal component of force F :<\/span><\/p>\n

Fx<\/sub> = F cos \u03b8 <\/span><\/p>\n

Fx<\/sub> = (30)(0.8) = 24 N <\/span><\/p>\n

The vertical component of force F :<\/span><\/p>\n

Fy<\/sub> = F sin \u03b8 <\/span><\/p>\n

Fy<\/sub> = (30)(0.6) = 18 N <\/span><\/p>\n

The normal force on object A :<\/span><\/p>\n

NA<\/sub> = wA <\/sub>= mA<\/sub> g = (4)(10) = 40 N<\/span><\/p>\n

The normal force on object B :<\/span><\/p>\n

NB<\/sub> + Fy<\/sub> \u2013 wB<\/sub> = 0<\/span><\/p>\n

NB<\/sub> + Fy<\/sub> = wB<\/sub><\/span><\/p>\n

NB<\/sub> = wB<\/sub> \u2013 Fy <\/sub>= mB<\/sub> g \u2013 Fy <\/sub>= (2)(10) \u2013 18 = 20 \u2013 18 = 2 N <\/span><\/p>\n

The force of kinetic friction between object A and floor :<\/span><\/p>\n

fkA<\/sub> = \u03bck <\/sub>NA<\/sub> = (0.2)(40) = 8 N<\/span><\/p>\n

The force of kinetic friction between object B and floor :<\/span><\/p>\n

fkB<\/sub> = \u03bck <\/sub>NB<\/sub> = (0.2)(2) = 0.4 N<\/span><\/p>\n

Acceleration of both objects :<\/b><\/span><\/p>\n

\u03a3F = m a<\/span><\/p>\n

Fx<\/sub> – fkB<\/sub> \u2013 fkA<\/sub> = (mA<\/sub> + mB<\/sub>) a <\/span><\/p>\n

24 \u2013 0.4 \u2013 8 = (4 + 2) a<\/span><\/p>\n

15.6 = 6 a<\/span><\/p>\n

a = 15.6 \/ 6<\/span><\/p>\n

a = 2.6 m\/s2<\/sup><\/span><\/p>\n

3. Two objects connected by a cord over a pulley as shown in figure below. Mass of object A = mA<\/sub>, mass f object B = mB <\/sub>and the acceleration of block B is a. Acceleration due to gravity is g. What is the tension force on block B.<\/span><\/p>\n

Solution :\"Dynamics<\/span><\/p>\n

The horizontal surface is smooth so there is no friction. The only force that accelerates system is the weight of block B.<\/span><\/p>\n

System’s acceleration :<\/u><\/span><\/p>\n

\"Dynamics<\/span><\/p>\n

The tension force (T) :<\/u><\/span><\/p>\n

\"Dynamics<\/span><\/p>\n

Substitute mA<\/sub> in equation 1 with mA <\/sub>in equation 2.<\/span><\/p>\n

\"Dynamics<\/span><\/p>\n

    \n
  1. What is the difference between statics and dynamics in the context of particles?<\/strong><\/span>\n