Kötélekkel és csigákkal összekötött testek egyensúlya – Newton első főtételének problémái és megoldásai

1. A box of tömeg 5 kg is on an inclined plane at an angle 30o. The box supported by a cord. Determine the tension force (T) and the normál erő (N)!

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 1

Megoldás

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 2ΣFx = 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

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2. Two objects of mass m1 = m2 = 2 kg, connected by a massless string over a frictionless pulley. Find the tension force T1 és T2.

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 3

Megoldás

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 4

(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

Jelentkezem Newton első törvénye to object 2 :

ΣFy = 0

T2 - w2 = 0

T2 = w2 = m2 g = (2 kg)(9.8 m/s2) = 19.6 N

T1 = T.2 = 19.6 N.

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3. An object of súly 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 statikus súrlódás between wB and inclined surface, if the system is at rest.

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 5

Megoldás

Equilibrium of bodies connected by cord and pulley – application of Newton's first law problems and solutions 6

(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

É – nyugatB 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 bűn 45o – T = 0

μs N + wB bűn 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.

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  1. Egydimenziós egyensúlyban lévő részecskék
  2. Részecskék kétdimenziós egyensúlyban
  3. Kötélekkel és csigákkal összekötött testek egyensúlya
  4. Equilibrium of bodies on inclined plane

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