imported>Patrick at 03:03, 13 June 2011

2011-06-13T03:03:09Z

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'''(a)''' the blocks are at rest, that is <math>a = 0</math>.
 
To find the value of <math>F_{T_1}</math>, we look at the two forces, <math>F_{T_1}</math> and <math>F_G</math>, acting on the system comprised of both masses, <math>m_A</math> and <math>m_B</math>, as shown: 
[[image:N_APP_EX_18_SOLNa.png|TOP]] 
<math>F_{T_1} - 5 N = 0</math> 
<math>F_{T_1} = 5 N</math>
 
To find the value of <math>F_{T_2}</math>, we isolate <math>m_B</math>: 
[[image:N_APP_EX_18_SOLNb.png|TOP]] 
<math>F_{T_2} - 3 N = 0</math> 
<math>F_{T_2} = 3 N</math>
 
'''(b)''' the blocks are moving at '''constant velocity, that is <math>a = 0</math>!'''. <math>F_{T_1} = 5 N</math> and <math>F_{T_2} = 3 N</math> as above. 

'''(c)''' The blocks accelerate upwards at <math>2 m/s^2</math>, that is, they have an acceleration of <math>+ 2 m/s^2</math>.
 
To find the value of <math>F_{T_1}</math>, we look at the forces acting at the system comprised of both masses, <math>m_A</math> and <math>m_B</math>, as shown in part a): 
[[image:N_APP_EX_18_SOLNc.png|TOP]] 
<math>F_{T_1} - 5 N = (m_A + m_B)a</math> 
<math>F_{T_1} - 5 N = (0.5 kg)(2 m/s^2)</math> 
<math>F_{T_1} = 5 N + 1 N = 6 N</math>
 
To find the value of <math>F_{T_2}</math>, we isolate <math>m_B</math> and look at the forces acting on it: 
[[image:N_APP_EX_18_SOLNd.png|TOP]] 
<math>F_{T_2} - 3 N = m_B a</math> 
<math>F_{T_2} - 3 N = (0.3 kg)(2 m/s^2)</math> 
<math>F_{T_2} = 3 N + 0.6 N = 3.6 N</math>
 
'''(d)''' The blocks accelerate downwards at <math>2 m/s^2</math>, that is they have an acceleration of <math>- 2 m/s^2</math>. Similarly to part c), we can find the value of <math>F_{T_1}</math> by looking at the forces acting at the system comprised of both masses, <math>m_A</math> and <math>m_B</math>: 
[[image:N_APP_EX_18_SOLNe.png|TOP]] 
<math>F_{T_1} - 5 N = (m_A + m_B)a</math> 
<math>F_{T_1} - 5 N = (0.5 kg)(-2 m/s^2)</math> 
<math>F_{T_1} = 5 N - 1 N = 4 N</math>
 
To find the value of <math>F_{T_2}</math>, we isolate <math>m_B</math> and look at the forces acting on it: 
[[image:N_APP_EX_18_SOLNf.png|TOP]] 
<math>F_{T_2} - 3 N = m_B a</math> 
<math>F_{T_2} - 3 N = (0.3 kg)(-2 m/s^2)</math> 
<math>F_{T_2} = 3 N - 0.6 N = 2.4 N</math>

Newtons Laws EX 18 - Revision history

imported>Patrick at 03:03, 13 June 2011