Tinthu tating'onoting'ono tomwe tili mu umodzi - kugwiritsa ntchito mavuto ndi mayankho a lamulo loyamba la Newton

1. Misa of an object, m = 10 kg, supported by a cord. Find the tension in the cord! g = 10 m/s2

Particles in one-dimensional equilibrium – application of Newton's first law problems and solutions 1Zodziwika:

Kulemera (m) = 10 kg

Kuthamanga chifukwa cha mphamvu yokoka (g) = 10 m/s2

Akufuna: Mphamvu ya kukakamira (T)

Yankho:

ΣFy = 0

T – w = 0

T = w

T = m g

T = (10 kg)(10 m/s2) = 100 kg m/s2

T = 100 Newtons

Onaninso  Kuchulukana ndi kuyandama mofanana - mavuto ndi mayankho

2. Mass of the object is 10 kg. Find the tension in the cord….. Acceleration due to gravity = 10 m/s2.

Anakonza

Zodziwika:

Kulemera (m) = 10 kg

Kuthamanga chifukwa cha mphamvu yokoka (g) = 10 m/s2.

Akufuna: Mphamvu ya kukakamira (T)

Yankho:

Particles in one-dimensional equilibrium – application of Newton's first law problems and solutions 2w = kulemera = m g = (10 kg)(10 m/s2) = 100 kg m/s2

T1 = the tension force 1

T1x = the x-component of the tension force 1 = T1 45o = 0.7 T1

T1y = the y-component of the tension force 2 = T1 tchimo 45o = 0.7 T1

T2 = the tension force 2

T2x = the x-component of the tension force 2 = T2 45o = 0.7 T2

T2y = the y-component of the tension force 2 = T2 tchimo 45o = 0.7 T2

The equilibrium condition ΣF = 0.

y axis :

ΣFy = 0

T1y + T2y – w = 0

0.7T1 + 0.7T2 - 100 = 0

0.7T1 + 0.7T2 = 100 —– equation 1

x axis :

ΣFx = 0

T2x - T1x = 0

0.7T2 – 0.7T1 = 0

0.7T2 = 0.7t1

T2 = T1 —– equation 2

Determine magnitude of T1 :

0.7T1 + 0.7T1 = 100

1.4T1 = 100

T1 = 100/1.4

T1 = 71.4 Newton

T1 = T2 so T2 = 71.4 Newton

[wpdm_package id='486′]

  1. Tinthu tating'onoting'ono tofanana
  2. Tinthu tating'onoting'ono tofanana mbali ziwiri
  3. Kufanana kwa matupi olumikizidwa ndi zingwe ndi ma pulley
  4. Kufanana kwa matupi omwe ali pamtunda wopendekeka

Siyani Comment