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 cuimilte cinéiteach idir m2 and incline is 0.1.
(a) Determine their luasghéarú
(b) Determine the tension force
Ar a dtugtar:
Aifreann 1 (m1) = Kg 2
Mais 2 (m2) = Kg 4
Coefficient of the kinetic friction between m1 agus eitleán claonta (μk1) = 0.2
Coefficient of the kinetic friction between m2 and inclined plane (μk2) = 0.1
Luasghéarú de bharr domhantarraingthe (g) = 9.8 m/s2
a) The magnitude and direction of the acceleration
w1 = meáchan 1 = m1 g = (2 kg)(9.8 m/s2) = 19.6 Niútón
w1x = w1 peaca 30o = (19.6 N)(0.5) = 9.8 Niútón
w1y = w1 cos 30o = (19.6 N)(0.87) = 17 Niútón
N1 = Tá an gnáthfhórsa phós muid1 = w1y = 17 Niútón
Fk1 = The force of the kinetic friction on m1 = μk1 N1 = (0.2)(17 N) = 3.4 Niútón
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w2 = weight 2 = m2 g = (4 kg)(9.8 m/s2) = 39.2 Niútón
w2x = w2 peaca 60o = (39.2 N)(0.87) = 34.1 Niútón
w2y = w2 cos 60o = (39.2 N)(0.5) = 19.6 Niútón
N2 = The normal force on m2 = w2y = 19.6 Niútón
Fk2 = The force of the kinetic friction on m2 = μk2 N2 = (0.1)(19.6 N) = 1.96 Niútón
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Méid an luasghéaraithe:
∑Fx = máx
w2x > w1x so direction of the acceleration is the same as direction of w2x.
Forces which points along acceleration is positive and forces which has opposite direction with acceleration is negative.
w2x - F.k2 - T.2 + T.1 - w1x - F.k1 = (m1 +m2) Anx
w2x - F.k2 - w1x - F.k1 = (m1 +m2 ) Anx
34.1 N – 1.96 N – 9.8 N – 3.4 N = (2 kg + 4 kg) ax
18.94 N = (6 kg) ax
ax = 18.94 N : 6 kg
ax = 3.16 m/s2
Méid an luasghéaraithe = 3.16 m/s2 . Direction of the acceleration = direction of T1 = direction of w2x
b) Magnitude of the tension force
Apply Newton’s second law on the object 2 :
w2x - F.k2 - T.2 = m2 ax
34.1 N – 1.96 N – T2 = (4kg)(3.16m/s2)
32.14 N – T2 = 12.64 N.
T2 = 32.14 N – 12.64 N = 19.5 Newton
The tension force = T = T1 = T.2 = 19.5 Niútón
2. m1 = 4 kg, m2 = 2 kg. Determine (a) magnitude and direction of the acceleration (b) Magnitude of the tension force which connecting m1 agus m2 (c) magnitude of the tension force which connecting pulley and roof.
réiteach
w1 = m1 g = (4 kg)(9.8 m/s2) = 39.2 Niútón
w2 = m2 g = (2 kg)(9.8 m/s2) = 19.6 Niútón
a) Magnitude and direction of the acceleration
∑Fy = máy
w1 > w2 so 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 - T.1 + T.2 - w2 = (m1 +m2) Any
w1 - w2 = (m1 +m2) Any
39.2 N – 19.6 N = (4 kg + 2 kg) ay
19.6 N = (6 kg) ay
ay = 19.6 N : 6 kg
ay = 3.26 m/s2
Magnitude of acceleration = 3.26 m/s2. Direction of acceleration = direction of w1 .
b) Magnitude of tension force which connecting m1 agus m2
Cuir iarratas isteach An dara dlí ag Newton phós muid2 :
∑Fy = máy
w1 - T.1 = m1 ay
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 Niútón
Magnitude of the tension force which connection objects = T = T1 = T.2 = 26.16 Niútón
c) Magnitude of the tension force which connecting pulley and roof.
Pulley is at rest :
∑Fy = máy —— ay = 0
∑Fy = 0
Upward force are positive, downward forces are negative :
T3 - T.1 - T.2 = 0
T3 = T.1 + T.2
T1 agus T2 have the same magnitude, T1 = T.2 = T = 26.16 N :
T3 = 2T = 2(26.16 N) = 52.32 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.
réiteach
w1 = The weight of the block 1 = m1 g = (10 kg)(9.8 m/s2) = 98 Niútón
w2 = The weight of the block 2 = m2 g = (15 kg)(9.8 m/s2) = 147 Niútón
w2y = w2 cos 30o = (147 N)(0.87) = 127.89 Niútón
w2x = w2 peaca 30o = (147 N)(0.5) = 73.5 Niútón
N2 = The normal force on the block 2 = w2y = 127.89 Niútón
Fk2 = The force of the kinetic friction on the block 2 = μk2 N2 = (0.42)(127.89 N) = 53.7 Niútón
Fs2 = The force of the static friction on the block 2 = μs2 N2 = (0.6)(127.89 N) = 76.7 Niútón
a) The magnitude of the minimum force F exerted on the objects so the objects accelerated upward
∑Fx = máx —— ax = 0
∑Fx = 0
Upward forces and rightward forces are positive, downward forces and leftward forces are negative.
F – Fk2 - w2x - w1 - T.2 + T.1 = 0
F – Fk2 - w2x - w1 = 0
F = Fk2 +w2x +w1
F = 53.7 N + 73.5 N + 98 N
F = 225.2 Niútón
b) The magnitude of the tension force
Apply Newton’s law of the motion on the block 1 :
∑Fy = máy —— ay = 0
∑Fy = 0
T1 - w1 = 0
T1 = w1 = 98 Niútón
Apply Newton’s law of the motion on the block 2 :
F – Fk2 - w2x - T.2 = 0
T2 = F – Fk2 - w2x
T2 = 225.2 N – 53.7 N – 73.5 N
T2 = 98 Niútón
Méid an fhórsa teannais = T1 = T.2 = T = 98 Niútón
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 ?
Réiteach:
a) When the system is released from rest, the block 3 and the block 2 still slide together?
w1 = Tá an weight of the block 1 = m1 g = (16 kg)(9.8 m/s2) = 156.8 Niútón
w1x = w1 peaca 60o = (156.8 N)(0.87) = 136.4 Niútón
w1y = w1 cos 60o = (156.8 N)(0.5) = 78.4 Niútón
N1 = Tá an normal force exerted on the block 1 by the inclined plane = w1y = 78.4 Niútón
w3 = Tá an weight of the block 3 = m3 g = (5 kg)(9.8 m/s2) = 49 Niútón
N23 = Tá an normal force exerted on the block 3 bythe block 2 = w3 = 49 Niútón
N32 = The normal force exerted on the block 2 by the block 3 = N23 = w3 = 49 Niútón
(N23 agus N32 are action-reaction pair)
Fs23 = Tá an force of the static friction exerted on the block 3 by the block 2 = μs N23 = (0.3)(49 N) = 14.7 newton
Fs32 = Tá an force of the static friction exerted on th block 2 by the block 3 = F.s23 = 14.7 Niútón
(Fs23 agus Fs32 are action-reaction pair)
w2 = Tá an weight of the block 2 = m2 g = (12 kg)(9.8 m/s2) = 117.6 Niútón
N2 = Tá an normal force exerted on the object 2 by the horizontal surface = w2 + N.32 = 117.6 Newton + 49
Newton = 166.6 Newton
Fk2 = Tá an force of the kinetic friction on the block 2 = μk N2 = (0.4)(166.6 N) = 66.64 Niútón
Apply Newton’s law of motion on the block 3 :
∑Fx = máx
Fs23 =m3 ax
—–> Fs23 = μs N23 = μs w3 = μs m3 g
μs m3 g = m3 ax
μs g = ax
ax = (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 w1x.
∑Fx = máx
w1x - T.1 + T.2 - F.k2 - F.s32 + Fs23 = (m1 +m2 +m3) Anx
w1x - F.k2 = (m1 +m2 +m3 ) Anx
136.4 N – 66.64 N = (16 kg + 12 kg + 5 kg) ax
69.76 N = (33 kg) ax
ax = 2.11 m/s2
ax is positive, means direction of the block displacement or the direction of the acceleration is same as direction of T2 or direction of w1x.
The magnitude of the acceleration is 2.11 m / s2 , anower than 2.94 m / s2 so 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
∑Fx = máx
w1x - F.k2 = (m1 +m2) Anx
—–> Fk2 = μk N2 = μk w2 = μk m2 g = (0.4)(12 kg)(9.8 m/s2) = 47.04 Niútón
136.4 N – 47.04 N = (16 kg + 12 kg) ax
89.36 N = (28 kg) ax
ax = 89.36 N : 28 kg = 3.19 m/s2
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- Mais agus meáchan
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