1. A box of masė 5 kg is on an inclined plane at an angle 30o. The box supported by a cord. Determine the tension force (T) and the normali jėga (N)!
Sprendimas
∑Fx = 0
T – w sin 30o = 0
T = w sin 30o
T = (5 kg)(9.8 m/s2) sin 30o
T = (49)(0.5)
T = 24.5 niutonų
∑Fy = 0
N – w cos 30o = 0
N = w cos 30o
N = (49)(0.87)
N = 43 niutono
2. Two objects of mass m1 = m2 = 2 kg, connected by a massless string over a frictionless pulley. Find the tension force T1 ir t2.
Sprendimas
(a) Free-body diagram for object 1 (b) Free-body diagram for object 2
Apply Newton’s first law to object 1 :
∑Fy = 0
T1 - w1 = 0
T1 = w1 = m1 g = (2 kg)(9.8 m/s2) = 19.6 N
Aplikuoti Pirmasis Niutono dėsnis to object 2 :
∑Fy = 0
T2 - w2 = 0
T2 = w2 = m2 g = (2 kg)(9.8 m/s2) = 19.6 N
T1 = T.2 = 19.6 N.
3. An object of svoris wA = 30 N and an object of weight wB = 40 N, are attached by a lightweight cord that passes over a frictionless pulley of the negligible mass. Determine the coefficient of the maximum statinė trintis between wB and inclined surface, if the system is at rest.
Sprendimas
(a) Free-body diagram for object wA (b) Free-body diagram for object wB
Apply Newton’s first law to object wA in vertical (y) direction :
∑Fy = 0 (no acceleration in vertical direction)
T – wA = 0
T = wA = 30 niutonų
Apply Newton’s first law to object wB in vertical (y) direction :
∑Fy = 0
Š – vB cos 45o = 0
N = wB cos 45o = (40)(0.7) = 28 Niutono
Apply Newton’s first law to object wB in horizontal (x) direction :
∑Fx = 0
Fk +wB 45. nuodėmėo – T = 0
μs N + wB 45. nuodėmėo – T = 0
μs (28) + (40)(0.7) – 30 = 0
μs (28) + 28 – 30 = 0
μs (28) = 30 – 28
μs (28) = 2
μs = 2/28
μs = 0.07
The coefficient of the maximum static friction between wB and inclined surface = 0.07.
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