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

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

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

Solution

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

Fy = 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 and T2.

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

Solution

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 :

Fy = 0

T1 – w1 = 0

T1 = w1 = m1 g = (2 kg)(9.8 m/s2) = 19.6 N

Apply Newton’s first law 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.

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3. An object of weight 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 static friction 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

Solution

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 :

Fy = 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 :

Fy = 0

N – wB cos 45o = 0

N = wB cos 45o = (40)(0.7) = 28 Newton

Apply Newton’s first law to object wB in horizontal (x) direction :

Fx = 0

Fk + wB sin 45o – T = 0

μs N + wB sin 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.

  1. Particles in one-dimensional equilibrium
  2. Particles in two-dimensional equilibrium
  3. Equilibrium of bodies connected by cord and pulley
  4. Equilibrium of bodies on inclined plane