imported>Patrick at 18:33, 26 May 2011

2011-05-26T18:33:09Z

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Since the bob is held in place, its acceleration is equal to zero. There are three forces acting on the bob: the force exerted by the spring <math>F_S</math>, the force exerted by the string <math>F_T</math> and the force of gravity <math>F_G</math>. Assume that the mass of the bob is <math>m</math> and then <math>F_G = m g</math>. We choose the coordinate system with the horizontal x-axis and draw the free body diagram. Given that the string makes a 30° angle with the vertical, the vector <math>F_T</math> makes a 60° angle with the x-axis. 
[[image:N_APP_EX_26_SOLN.png|TOP]] 

Next we compute the components of all forces. We begin by computing the magnitude of <math>\overrightarrow{F_S}</math>:

<math>F_S = k x = (3.92 \times 10^3)(2 \times 10^{-2}) = 78.4 N</math>
 
{| border="1" cellpadding="2" style="text-align:center"
|-
! Forces
! x-component
! y-component
|-
! <math>F_S</math>
| -78.4 N
| 0
|-
! <math>F_G</math>
| 0
| -mg
|-
! <math>F_T</math>
| <math>F_T \cos(60^{\circ})</math>
| <math>F_T \sin(60^{\circ})</math>
|-
! <math>\Sigma F</math>
| 0
| 0
|}
 
The sum of the forces in the x-direction is zero:

<math>-78.4 + F_T \cos(60^{\circ}) = 0</math>

<math>F_T = {78.4 \over \cos(60^{\circ})} = {78.4 \over 0.5} = 157 N</math> 

The sum of the forces in the y-direction is zero:

<math>F_T \sin(60^{\circ}) - m g = 0</math>

<math>m = {F_T \sin(60^{\circ}) \over g} = {0.87 F_T \over g} = {0.87 (157) \over (10)} = 13.7 kg</math>

Newtons Laws EX 26 - Revision history

imported>Patrick at 18:33, 26 May 2011