Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions

1. Two masses m₁= 2 kg me te 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 te waku nekeneke between m₂ and incline is 0.1.

(a) Determine their whakatere

(b) Determine the tension force

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 1

Mōhiotia:

Mass 1 (m₁) = 2 kirokaramu

Taumaha 2 (m₂) = 4 kirokaramu

Coefficient of the kinetic friction between m₁ a papa whakarara (μ_k1) = 0.2

Coefficient of the kinetic friction between m₂ and inclined plane (μ_k2) = 0.1

Whakaterenga na te kaha o te kaha (g) = 9.8 m/s²

a) The magnitude and direction of the acceleration

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 2

w₁ = taimaha 1 = m₁karamu = (2 kg)(9.8 m/s²) = 19.6 Niutona

w_1x =w₁ hara 30^o = (19.6 N)(0.5) = 9.8 Nīutona

w_1y =w₁ whaimana 30^o= (19.6 N)(0.87) = 17 Nīutona

N₁= The kaha noa i runga i te m₁ =w_1y = 17 Niutona

F_k1 = The force of the kinetic friction on m₁ = μ_k1N₁ = (0.2)(17 N) = 3.4 Nīhana

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w₂ = weight 2 = m₂karamu = (4 kg)(9.8 m/s²) = 39.2 Niutona

w_2x =w₂hara 60^o = (39.2 N)(0.87) = 34.1 Nīutona

w_2y =w₂ whaimana 60^o= (39.2 N)(0.5) = 19.6 Nīutona

N₂ = The normal force on m₂ =w_2y= 19.6 Niutona

F_k2= The force of the kinetic friction on m₂ = μ_k2 N₂ = (0.1)(19.6 N) = 1.96 Nīhana

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Te rahi o te whakaterenga:

∑F_x = mā_x

w_2x> w_1x so direction of the acceleration is the same as direction of w_2x.

Forces which points along acceleration is positive and forces which has opposite direction with acceleration is negative.

w_2x- F_k2 - T₂ +T₁ - w_1x - F_k1= (m₁ +m₂) a_x

w_2x - F_k2 - w_1x - F_k1 = (m₁ +m₂ ) a_x

34.1 N – 1.96 N – 9.8 N – 3.4 N = (2 kg + 4 kg) a_x

18.94 N = (6 kg) a_x

a_x = 18.94 N : 6 kg

a_x= 3.16m/s²

Te rahi o te whakaterenga = 3.16 m/s² . Direction of the acceleration = direction of T₁= direction of w_2x

b) Magnitude of the tension force

Apply Newton’s second law on the object 2 :

w_2x- F_k2- T₂ = m₂a_x

34.1 N – 1.96 N – T₂= (4 kg)(3.16 m/s²)

32.14 N – T₂ = 12.64 N

T₂= 32.14 N – 12.64 N = 19.5 Newton

The tension force = T = T₁ =T₂ = 19.5 Niutona

A hi'o atoa Ngā ngaru miihini (Auau Wā Roangaru Te tere o te ngaru) - Ngā Raru me ngā Otinga

2. m₁= 4 kg, mita₂ = 2 kg. Determine (a) magnitude and direction of the acceleration (b) Magnitude of the tension force which connecting m₁ me te m₂ (c) magnitude of the tension force which connecting pulley and roof.

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 3

otinga

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 4

w₁ = m₁karamu = (4 kg)(9.8 m/s²) = 39.2 Niutona

w₂= m₂ karamu = (2 kg)(9.8 m/s²) = 19.6 Niutona

a) Magnitude and direction of the acceleration

∑F_y = mā_y

w₁> w₂so the direction of the object is same as the direction of the weight 1 (w₁). Forces which has the same direction with acceleration is positive and forces which has opposite direction with acceleration is negative.

w₁- T₁ +T₂ - w₂= (m₁ +m₂) a_y

w₁ - w₂ = (m₁ +m₂) a_y

39.2 N – 19.6 N = (4 kg + 2 kg) a_y

19.6 N = (6 kg) a_y

a_y= 19.6 N : 6 kg

a_y = 3.26m/s²

Magnitude of acceleration = 3.26 m/s². Direction of acceleration = direction of w₁ .

b) Magnitude of tension force which connecting m₁ me te m₂

Anga Te ture tuarua a Newton i runga i te m₂ :

∑F_y = mā_y

w₁ - T₁ = m₁a_y

39.2 N – T₁= (4 kg)( 3.26 m/s²)

39.2 N – T₁ = 13.04 N

T₁= 39.2 N – 13.04 N

T₁ = 26.16 Niutona

Magnitude of the tension force which connection objects = T = T₁ =T₂ = 26.16 Niutona

c) Magnitude of the tension force which connecting pulley and roof.

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 5 Pulley is at rest :

∑F_y= mā_y —— a_y= 0

∑F_y= 0

Upward force are positive, downward forces are negative :

T₃ - T₁ - T₂ = 0

T₃=T₁ +T₂

T₁ a T₂ have the same magnitude, T₁ =T₂ = T = 26.16 N :

T₃ = 2T = 2(26.16 N) = 52.32 Newton

A hi'o atoa Te nekehanga i runga i te papa angiangi me te kaha waku - te whakamahinga o ngā raruraru me ngā otinga o te ture nekehanga a Newton

3. Block 1 (m₁ = 10 kg) and block 2 (m₂= 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.

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 6

otinga

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 7

w₁ = The weight of the block 1 = m₁ karamu = (10 kg)(9.8 m/s²) = 98 Niutona

w₂ = The weight of the block 2 = m₂ karamu = (15 kg)(9.8 m/s²) = 147 Niutona

w_2y =w₂ whaimana 30^o = (147 N)(0.87) = 127.89 Nīutona

w_2x =w₂hara 30^o = (147 N)(0.5) = 73.5 Nīutona

N₂ = The normal force on the block 2 = w_2y= 127.89 Niutona

F_k2 = The force of the kinetic friction on the block 2 = μ_k2 N₂ = (0.42)(127.89 N) = 53.7 Nīhana

F_s2 = The force of the static friction on the block 2 = μ_s2 N₂ = (0.6)(127.89 N) = 76.7 Nīhana

a) The magnitude of the minimum force F exerted on the objects so the objects accelerated upward

∑F_x = mā_x —— a_x = 0

∑F_x = 0

Upward forces and rightward forces are positive, downward forces and leftward forces are negative.

Wh – Wh_k2 - w_2x - w₁ - T₂ +T₁ = 0

Wh – Wh_k2 - w_2x- w₁ = 0

F = F_k2+ w_2x + w₁

F = 53.7 N + 73.5 N + 98 N

F = 225.2 Newton

b) The magnitude of the tension force

Apply Newton’s law of the motion on the block 1 :

∑F_y = mā_y—— a_y= 0

∑F_y = 0

T₁ - w₁ = 0

T₁ =w₁ = 98 Niutona

Apply Newton’s law of the motion on the block 2 :

Wh – Wh_k2 - w_2x - T₂ = 0

T₂= F – F_k2- w_2x

T₂= 225.2 N – 53.7 N – 73.5 N

T₂ = 98 Niutona

Te rahi o te kaha kume = T₁ =T₂ = T = 98 Niutona

A hi'o atoa Te kaha pūrua – ngā raruraru me ngā otinga

4. Block 1 (m₁ = 16 kg) lies on a horizontal surface and the block 2 (m₂ = 12 kg) lies on a smooth inclined plane, connected by a cord that passes over a small, frictionless pulley. Block 3 (m₃ = 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 coefficient of the static friction between the block 2 with the block 3 is 0,3.

(a) When the system is released from rest, the block 3 and the block 2 still slide together ?

(B) If there is no block 3, what is the acceleration of the block 1 and the block 2 ?

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 8

Rongoā:

a) When the system is released from rest, the block 3 and the block 2 still slide together?

Two bodies with the same magnitude of acceleration – Application of Newton's law of motion problems and solutions 9

w₁ = The weight of the block 1 = m₁ karamu = (16 kg)(9.8 m/s²) = 156.8 Niutona

w_1x =w₁ hara 60^o = (156.8 N)(0.87) = 136.4 Nīutona

w_1y =w₁ whaimana 60^o = (156.8 N)(0.5) = 78.4 Nīutona

N₁ = The normal force exerted on the block 1 by the inclined plane =w_1y = 78.4 Niutona

w₃ = The weight of the block 3 = m₃ karamu = (5 kg)(9.8 m/s²) = 49 Niutona

N₂₃ = The normal force exerted on the block 3 bythe block 2 =w₃= 49 Niutona

N₃₂ = The normal force exerted on the block 2 by the block 3 = N₂₃ =w₃ = 49 Niutona

(N₂₃a N₃₂ are action-reaction pair)

F_s23 = The force of the static friction exerted on the block 3 by the block 2 = μ_s N₂₃= (0.3)(49 N) = 14.7 Newton

F_s32= The force of the static friction exerted on th block 2 by the block 3 =F_s₂₃ = 14.7 Niutona

(F_s23 a F_s32 are action-reaction pair)

w₂ = The weight of the block 2 = m₂ karamu = (12 kg)(9.8 m/s²) = 117.6 Niutona

N₂ = The normal force exerted on the object 2 by the horizontal surface =w₂ +N₃₂ = 117.6 Newton + 49

Newton = 166.6 Newton

F_k2= The force of the kinetic friction on the block 2 = μ_k N₂ = (0.4)(166.6 N) = 66.64 Nīhana

Apply Newton’s law of motion on the block 3 :

∑F_x = mā_x

F_s23=m₃a_x

—–> F_s23 = μ_sN₂₃= μ_sw₃= μ_sm₃g

μ_s m₃g = m₃a_x

μ_s g = a_x

a_x = (0.3)(9.8 m/s²) = 2.94 m/s²

The maximum acceleration of the block 3 so that the block 3 and the block 2 still slide together is 2.94 m/s².

Now we calculate the magnitude of the system’s acceleration after released from rest.

The direction of the block displacement = the direction of the block’s acceleration = the direction of T₂= the direction of w_1x.

∑F_x = mā_x

w_1x - T₁+T₂- F_k2 - F_s32 +F_s23 = (m₁ +m₂ +m₃) a_x

w_1x - F_k2 = (m₁ +m₂+m₃ ) a_x

136.4 N – 66.64 N = (16 kg + 12 kg + 5 kg) a_x

69.76 N = (33 kg) a_x

a_x = 2.11m/s²

a_x is positive, means direction of the block displacement or the direction of the acceleration is same as direction of T₂ or direction of w_1x.

The magnitude of the acceleration is 2.11 m / s² , lower than 2.94 m / s²so we can conclude that block 3 and block 2 still slide together after released from rest.

b) The magnitude of the acceleration of the block 1 and the block 2

∑F_x= mā_x

w_1x - F_k2= (m₁ +m₂) a_x

—–> F_k2 = μ_k N₂= μ_k w₂ = μ_k m₂g = (0.4)(12 kg)(9.8 m/s²) = 47.04 Niutona

136.4 N – 47.04 N = (16 kg + 12 kg) a_x

89.36 N = (28 kg) a_x

a_x= 89.36 N : 28 kg = 3.19 m/s²

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Two bodies with the same magnitude of acceleration – Application of Newton’s law of motion problems and solutions

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