1. A box of papatipu 5 kg is on an inclined plane at an angle 30o. The box supported by a cord. Determine the tension force (T) and the kaha noa (N)!
otinga
∑Fx = 0
T – w sin 30o = 0
T = w sin 30o
T = (5 kg)(9.8 m/s)2) sin 30o
T = (49)(0.5)
T = 24.5 Ngā Niutona
∑Fy = 0
N – w cos 30o = 0
N = w cos 30o
N = (49)(0.87)
N = 43 Niutona
2. Two objects of mass m1 = m2 = 2 kg, connected by a massless string over a frictionless pulley. Find the tension force T1 a T2.
otinga
(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
Anga Te ture tuatahi a Newton to object 2 :
∑Fy = 0
T2 - w2 = 0
T2 =w2 = m2 g = (2 kg)(9.8 m/s2) = 19.6 N
T1 =T2 = 19.6 N.
3. An object of taimaha 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 te waku pumau between wB and inclined surface, if the system is at rest.
otinga
(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 Niutona
Apply Newton’s first law to object wB in vertical (y) direction :
∑Fy = 0
N – wB whaimana 45o = 0
N = wB whaimana 45o = (40)(0.7) = 28 Ngā Newton
Apply Newton’s first law to object wB in horizontal (x) direction :
∑Fx = 0
Fk + wB hara 45o – T = 0
μs N + wB hara 45o – 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.
[wpdm_package id='490′]
- Particles in one-dimensional equilibrium
- Particles in two-dimensional equilibrium
- Equilibrium of bodies connected by cord and pulley
- Equilibrium of bodies on inclined plane
