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 kinetika friction eo anelanelan'ny m2 and incline is 0.1.
(a) Determine their haingana
(b) Determine the tension force
Fantatra:
Lamesa 1 (m1) = 2 kg
Lanja 2 (m2) = 4 kg
Coefficient of the kinetic friction between m1 sy fiaramanidina mirona (μk1) = 0.2
Coefficient of the kinetic friction between m2 and inclined plane (μk2) = 0.1
Fanafainganana noho ny hery misintona (g) = 9.8 m/s2
a) The magnitude and direction of the acceleration
w1 = lanja 1 = m1 g = (2 kg)(9.8 m/s)2) = 19.6 Newton
w1x = w1 ota 30o = (19.6 N)(0.5) = 9.8 Newton
w1y = w1 ny 30o = (19.6 N)(0.87) = 17 Newton
N1 = Ny hery ara-dalàna on m1 = w1y = 17 Newton
Fk1 = The force of the kinetic friction on m1 = μk1 N1 = (0.2)(17 N) = 3.4 Newton
---
w2 = weight 2 = m2 g = (4 kg)(9.8 m/s)2) = 39.2 Newton
w2x = w2 ota 60o = (39.2 N)(0.87) = 34.1 Newton
w2y = w2 ny 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
---
Ny haben'ny hafainganam-pandeha:
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) aryx
w2x - Fk2 - w1x - Fk1 = (m1 +m2 ) aryx
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
Haben'ny hafainganana = 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/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 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 ary m2 (c) magnitude of the tension force which connecting pulley and roof.
vahaolana
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) aryy
w1 - w2 = (m1 +m2) aryy
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 ary m2
Ampiharo Lalàna faharoan'i Newton 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 ary T2 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.
vahaolana
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 ny 30o = (147 N)(0.87) = 127.89 Newton
w2x = w2 ota 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
Haben'ny hery fihenjanana = 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 ?
vahaolana:
a) When the system is released from rest, the block 3 and the block 2 still slide together?
w1 = Ny weight of the block 1 = m1 g = (16 kg)(9.8 m/s)2) = 156.8 Newton
w1x = w1 ota 60o = (156.8 N)(0.87) = 136.4 Newton
w1y = w1 ny 60o = (156.8 N)(0.5) = 78.4 Newton
N1 = Ny normal force exerted on the block 1 by the inclined plane = w1y = 78.4 Newton
w3 = Ny weight of the block 3 = m3 g = (5 kg)(9.8 m/s)2) = 49 Newton
N23 = Ny 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 sy N32 are action-reaction pair)
Fs23 = Ny force of the static friction exerted on the block 3 by the block 2 = μs N23 = (0.3)(49 N) = 14.7 Newton
Fs32 = Ny force of the static friction exerted on th block 2 by the block 3 =Fs23 = 14.7 Newton
(Fs23 sy Fs32 are action-reaction pair)
w2 = Ny weight of the block 2 = m2 g = (12 kg)(9.8 m/s)2) = 117.6 Newton
N2 = Ny normal force exerted on the object 2 by the horizontal surface = w2 +N32 = 117.6 Newton + 49
Newton = 166.6 Newton
Fk2 = Ny 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/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.
ΣFx = max
w1x - T1 +T2 - Fk2 - Fs32 +Fs23 = (m1 +m2 +m3) aryx
w1x - Fk2 = (m1 +m2 +m3 ) aryx
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 , lower 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) aryx
—–> 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′]
- Mass sy lanja
- Hery mahazatra
- Ny lalàna faharoa momba ny fihetsika nataon'i Newton
- Hery mikorontana
- Fihetsehana eo amin'ny velarana mitsivalana tsy misy hery mikoriana
- Ny fihetsehan'ny zavatra roa mitovy hafainganam-pandeha eo amin'ny velaran-tany marindrano marokoroko miaraka amin'ny herin'ny fikikisana
- Fihetsehana eo amin'ny sehatra mitongilana tsy misy hery mikorontana
- Fihetsehana eo amin'ny sehatra mitongilana mikitoantoana miaraka amin'ny herin'ny fikikisana
- Fihetsehana ao anaty ascenseur
- Ny fihetsehan'ny vatana dia mifandray amin'ny alalan'ny tady sy ny pulleys
- Vatana roa mitovy habe amin'ny hafainganana
- Famadihana fiolahana fisaka - dinamikan'ny fihetsehana boribory
- Famadihana fiolahana miolikolika - dinamikan'ny fihetsehana boribory
- Fihetsehana mitovy amin'ny faribolana mitsivalana
- Hery centripetal amin'ny fihetsehana boribory mitovy
