1. A box of massa 5 kg is on an inclined plane at an angle 30o. The box supported by a cord. Determine the tension force (T) and the forza normale (N)!
Soluzione
ΣFx = 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 Newton
ΣFy = 0
N – w cos 30o = 0
N = w cos 30o
N = (49)(0.87)
N = 43 Newton
2. Two objects of mass m1 = m2 = 2 kg, connected by a massless string over a frictionless pulley. Find the tension force T1 e T2.
Soluzione
(a) Free-body diagram for object 1 (b) Free-body diagram for object 2
Apply Newton’s first law to object 1 :
ΣFy = 0
T1 - w1 = 0
T1 = w1 = m1 g = (2 kg)(9.8 m/s2) = 19.6 N
APPLICA Prima legge di Newton to object 2 :
ΣFy = 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 peso 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 Frizione statica between wB and inclined surface, if the system is at rest.
Soluzione
(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 :
ΣFy = 0 (no acceleration in vertical direction)
T – wA = 0
T = wA = 30 Newton
Apply Newton’s first law to object wB in vertical (y) direction :
ΣFy = 0
N – oB cos 45o = 0
N = wB cos 45o = (40)(0.7) = 28 Newton
Apply Newton’s first law to object wB in horizontal (x) direction :
ΣFx = 0
Fk +wB peccato 45o – T = 0
μs N + wB peccato 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 ′]
- Particelle in equilibrio unidimensionale
- Particelle in equilibrio bidimensionale
- Equilibrio dei corpi collegati da corde e pulegge
- Equilibrio dei corpi su un piano inclinato
