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 动摩擦 between m2 and incline is 0.1.
(a) Determine their 促进
(b) Determine the tension force
已知：
质量 1（米）1) = 2公斤
Mass 2 (m2) = 4公斤
Coefficient of the kinetic friction between m1 和 斜面 （μk1) = 0.2
Coefficient of the kinetic friction between m2 and inclined plane (μk2) = 0.1
重力加速度 (g) = 9.8 米/秒2
a) The magnitude and direction of the acceleration
w1 = 重量 1 = m1 g = (2 kg)(9.8 m/s2) = 19.6 牛顿
w1x =w1 罪30o = (19.6 N)(0.5) = 9.8 牛顿
w1y =w1 cos 30o = (19.6 N)(0.87) = 17 牛顿
N1 =的 法向力 我们是1 =w1y = 17 牛顿
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/s2) = 39.2 牛顿
w2x =w2 罪60o = (39.2 N)(0.87) = 34.1 牛顿
w2y =w2 cos 60o = (39.2 N)(0.5) = 19.6 牛顿
N2 = The normal force on m2 =w2y = 19.6 牛顿
Fk2 = The force of the kinetic friction on m2 =μk2 N2 = (0.1)(19.6 N) = 1.96 Newton
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The magnitude of the acceleration :
∑Fx = 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 - T.2 + T1 -在1x - Fk1 =（米1 + 米2）的x
w2x - Fk2 -在1x - Fk1 =（米1 + 米2 ）的x
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 - T.2 =米2 ax
34.1 N – 1.96 N – T2 = (4 kg)(3.16 m/s2)
32.14 N – T2 = 12.64牛
T2 = 32.14 N – 12.64 N = 19.5 Newton
The tension force = T = T1 = T.2 = 19.5 牛顿
2.米1 = 4 kg，m2 = 2 kg. Determine (a) magnitude and direction of the acceleration (b) Magnitude of the tension force which connecting m1 和米2 (c) magnitude of the tension force which connecting pulley and roof.
解决方案
w1 =米1 g = (4 kg)(9.8 m/s2) = 39.2 牛顿
w2 =米2 g = (2 kg)(9.8 m/s2) = 19.6 牛顿
a) Magnitude and direction of the acceleration
∑Fy = 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 - T.1 + T2 -在2 =（米1 + 米2）的y
w1 -在2 =（米1 + 米2）的y
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 和米2
在断裂前， 牛顿第二定律 我们是2 :
∑Fy = may
w1 - T.1 =米1 ay
39.2 N – T1 = (4 kg)( 3.26 m/s2)
39.2 N – T1 = 13.04牛
T1 = 39.2 N – 13.04 N
T1 = 26.16 牛顿
Magnitude of the tension force which connection objects = T = T1 = T.2 = 26.16 牛顿
c) Magnitude of the tension force which connecting pulley and roof.
Pulley is at rest :
∑Fy = may —— ay = 0
∑Fy = 0
Upward force are positive, downward forces are negative :
T3 - T.1 - T.2 = 0
T3 = T.1 + T2
T1 和T2 have the same magnitudeŤ1 = 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.
解决方案
w1 = The weight of the block 1 = m1 g = (10 kg)(9.8 m/s2) = 98 牛顿
w2 = The weight of the block 2 = m2 g = (15 kg)(9.8 m/s2) = 147 牛顿
w2y =w2 cos 30o = (147 N)(0.87) = 127.89 牛顿
w2x =w2 罪30o = (147 N)(0.5) = 73.5 牛顿
N2 = The normal force on the block 2 = w2y = 127.89 牛顿
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
∑Fx = max —— ax = 0
∑Fx = 0
Upward forces and rightward forces are positive, downward forces and leftward forces are negative.
F – Fk2 -在2x -在1 - T.2 + T1 = 0
F – Fk2 -在2x -在1 = 0
F = Fk2 +w2x +w1
F = 53.7 N + 73.5 N + 98 N
F = 225.2 牛顿
b) The magnitude of the tension force
Apply Newton’s law of the motion on the block 1 :
∑Fy = may —— ay = 0
∑Fy = 0
T1 -在1 = 0
T1 =w1 = 98 牛顿
Apply Newton’s law of the motion on the block 2 :
F – Fk2 -在2x - T.2 = 0
T2 = F – Fk2 -在2x
T2 = 225.2 N – 53.7 N – 73.5 N
T2 = 98 牛顿
Magnitude of the tension force = T1 = T.2 T = 98 牛顿
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.
（一个） When the system is released from rest, the block 3 and the block 2 still slide together ?
（二） If there is no block 3, what is the acceleration of the block 1 and the block 2 ?
解决方案：
a) When the system is released from rest, the block 3 and the block 2 still slide together?
w1 =的 weight of the block 1 = m1 g = (16 kg)(9.8 m/s2) = 156.8 牛顿
w1x =w1 罪60o = (156.8 N)(0.87) = 136.4 牛顿
w1y =w1 cos 60o = (156.8 N)(0.5) = 78.4 牛顿
N1 =的 normal force exerted on the block 1 by the inclined plane =w1y = 78.4 牛顿
w3 =的 weight of the block 3 = m3 g = (5 kg)(9.8 m/s2) = 49 牛顿
N23 =的 normal force exerted on the block 3 bythe block 2 =w3 = 49 牛顿
N32 = The normal force exerted on the block 2 by the block 3 = N.23 =w3 = 49 牛顿
(N23 和 N32 are action-reaction pair)
Fs23 =的 force of the static friction exerted on the block 3 by the block 2 =μs N23 = (0.3)(49 N) = 14.7 Newton
Fs32 =的 force of the static friction exerted on th block 2 by the block 3 =Fs23 = 14.7 牛顿
(Fs23 和 Fs32 are action-reaction pair)
w2 =的 weight of the block 2 =米2 g = (12 kg)(9.8 m/s2) = 117.6 牛顿
N2 =的 normal force exerted on the object 2 by the horizontal surface =w2 + N32 = 117.6 Newton + 49
Newton = 166.6 Newton
Fk2 =的 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 :
∑Fx = max
Fs23 =m3 ax
—–> Fs23 =μs N23 =μs w3 =μs m3 g
μs m3 克 = 米3 ax
μs g = ax
ax = (0.3)(9.8 m/s2) = 2.94 米/秒2
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 = max
w1x - T.1 + T2 - Fk2 - Fs32 + Fs23 =（米1 + 米2 + 米3）的x
w1x - Fk2 =（米1 + 米2 + 米3 ）的x
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米/秒2 中，ower than 2.94米/秒2 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 = max
w1x - Fk2 =（米1 + 米2）的x
—–> Fk2 =μk N2 =μk w2 =μk m2 g = (0.4)(12 kg)(9.8 m/s2) = 47.04 牛顿
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']
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