1. Two masses m1 = 2 kg sareng 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 gesekan kinétik between m2 and incline is 0.1.
(a) Determine their gagancangan
(b) Determine the tension force
Dipikanyaho:
massa 1 (m1) = 2 kg
Beurat 2 (m2) = 4 kg
Coefficient of the kinetic friction between m1 jeung bidang miring (μk1) = 0.2
Coefficient of the kinetic friction between m2 and inclined plane (μk2) = 0.1
Akselerasi alatan gravitasi (g) = 9.8 m/s2
a) The magnitude and direction of the acceleration
w1 = beurat 1 = m1 g = (2 kg)(9.8 m/s)2) = 19.6 Newton
w1x = w1 dosa 30o = (19.6 N)(0.5) = 9.8 Newton
w1y = w1 éta 30o = (19.6 N)(0.87) = 17 Newton
N1 = The gaya normal on m1 = w1y = 17 Newton
Fk1 = The force of the kinetic friction on m1 = μk1 N1 = (0.2)(17 N) = 3.4 Newton
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w2 = weight 2 = m2 g = (4 kg)(9.8 m/s)2) = 39.2 Newton
w2x = w2 dosa 60o = (39.2 N)(0.87) = 34.1 Newton
w2y = w2 éta 60o = (39.2 N)(0.5) = 19.6 Newton
N2 = The normal force on m2 = w2y = 19.6 Newton
Fk2 = The force of the kinetic friction on m2 = μk2 N2 = (0.1)(19.6 N) = 1.96 Newton
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Besarna akselerasi:
ΣFx = max
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 - Fk2 - T2 +T1 - w1x - Fk1 = (m1 +m2) jeungx
w2x - Fk2 - w1x - Fk1 = (m1 +m2 ) jeungx
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
Magnitude of the acceleration = 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 - Fk2 - T2 = m2 ax
34.1 N – 1.96 N – T2 = (4 kg)(3.16 m/s)2)
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 Newton
2. m1 = 4 kg, m2 = 2 kg. Determine (a) magnitude and direction of the acceleration (b) Magnitude of the tension force which connecting m1 sareng m2 (c) magnitude of the tension force which connecting pulley and roof.
leyuran
w1 = m1 g = (4 kg)(9.8 m/s)2) = 39.2 Newton
w2 = m2 g = (2 kg)(9.8 m/s)2) = 19.6 Newton
a) Magnitude and direction of the acceleration
ΣFy = may
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 - T1 +T2 - w2 = (m1 +m2) jeungy
w1 - w2 = (m1 +m2) jeungy
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 sareng m2
ngalamar Newton’s second law on m2 :
ΣFy = may
w1 - T1 = 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 Newton
Magnitude of the tension force which connection objects = T = T1 = T2 = 26.16 Newton
c) Magnitude of the tension force which connecting pulley and roof.
Pulley is at rest :
ΣFy = may —— ay = 0
ΣFy = 0
Upward force are positive, downward forces are negative :
T3 - T1 - T2 = 0
T3 = T1 +T2
T1 jeung T.2 have the same magnitude, T1 = T2 = 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.
leyuran
w1 = The weight of the block 1 = m1 g = (10 kg)(9.8 m/s)2) = 98 Newton
w2 = The weight of the block 2 = m2 g = (15 kg)(9.8 m/s)2) = 147 Newton
w2y = w2 éta 30o = (147 N)(0.87) = 127.89 Newton
w2x = w2 dosa 30o = (147 N)(0.5) = 73.5 Newton
N2 = The normal force on the block 2 = w2y = 127.89 Newton
Fk2 = The force of the kinetic friction on the block 2 = μk2 N2 = (0.42)(127.89 N) = 53.7 Newton
Fs2 = The force of the static friction on the block 2 = μs2 N2 = (0.6)(127.89 N) = 76.7 Newton
a) The magnitude of the minimum force F exerted on the objects so the objects accelerated upward
ΣFx = max —— ax = 0
ΣFx = 0
Upward forces and rightward forces are positive, downward forces and leftward forces are negative.
F – Fk2 - w2x - w1 - T2 +T1 = 0
F – Fk2 - w2x - w1 = 0
F = Fk2 +w2x +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 :
ΣFy = may —— ay = 0
ΣFy = 0
T1 - w1 = 0
T1 = w1 = 98 Newton
Apply Newton’s law of the motion on the block 2 :
F – Fk2 - w2x - T2 = 0
T2 = F – Fk2 - w2x
T2 = 225.2 N – 53.7 N – 73.5 N
T2 = 98 Newton
Magnitude of the tension force = T1 = T2 = T = 98 Newton
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 ?
Solusi:
a) When the system is released from rest, the block 3 and the block 2 still slide together?
w1 = The weight of the block 1 = m1 g = (16 kg)(9.8 m/s)2) = 156.8 Newton
w1x = w1 dosa 60o = (156.8 N)(0.87) = 136.4 Newton
w1y = w1 éta 60o = (156.8 N)(0.5) = 78.4 Newton
N1 = The normal force exerted on the block 1 by the inclined plane = w1y = 78.4 Newton
w3 = The weight of the block 3 = m3 g = (5 kg)(9.8 m/s)2) = 49 Newton
N23 = The normal force exerted on the block 3 bythe block 2 = w3 = 49 Newton
N32 = The normal force exerted on the block 2 by the block 3 = N23 = w3 = 49 Newton
(N23 jeung N32 are action-reaction pair)
Fs23 = The force of the static friction exerted on the block 3 by the block 2 = μs N23 = (0.3)(49 N) = 14.7 Newton
Fs32 = The force of the static friction exerted on th block 2 by the block 3 =Fs23 = 14.7 Newton
(Fs23 jeung Fs32 are action-reaction pair)
w2 = The weight of the block 2 = m2 g = (12 kg)(9.8 m/s)2) = 117.6 Newton
N2 = The normal force exerted on the object 2 by the horizontal surface = w2 +N32 = 117.6 Newton + 49
Newton = 166.6 Newton
Fk2 = The force of the kinetic friction on the block 2 = μk N2 = (0.4)(166.6 N) = 66.64 Newton
Apply Newton’s law of motion on the block 3 :
ΣFx = max
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/dtk2
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.
ΣFx = max
w1x - T1 +T2 - Fk2 - Fs32 + Fs23 = (m1 +m2 +m3) jeungx
w1x - Fk2 = (m1 +m2 +m3 ) jeungx
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 , anuower 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
ΣFx = max
w1x - Fk2 = (m1 +m2) jeungx
—–> Fk2 = μk N2 = μk w2 = μk m2 g = (0.4)(12 kg)(9.8 m/s2) = 47.04 Newton
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
[wpdm_package id='493′]
- Massa sareng beurat
- Gaya normal
- Hukum gerak kadua Newton
- Gaya gesekan
- Gerakan dina permukaan horizontal tanpa gaya gesekan
- Gerakan dua awak kalayan akselerasi anu sami dina permukaan horizontal kasar kalayan gaya gesekan
- Gerak dina bidang miring tanpa gaya gesekan
- Gerak dina bidang miring kasar kalawan gaya gesekan
- Gerakan dina lift
- Gerakan awak disambungkeun ku tali sareng katrol
- Dua awak anu gaduh akselerasi anu sami
- Ngabulatkeun kurva datar - dinamika gerak sirkular
- Ngabulatkeun kurva miring - dinamika gerak sirkular
- Gerakan seragam dina bunderan horizontal
- Gaya séntripetal dina gerak sirkular seragam
