Question

1. below figure find tension on strings.

Explanation:

Step1: Analyze forces on each mass

For \(m_1 = 5\space kg\):
Normal force \(N_1=m_1g\cos70^{\circ}\), frictional force \(f_1=\mu_kN_1=\mu_km_1g\cos70^{\circ}\)
Force along the incline \(F_{1\parallel}=m_1g\sin70^{\circ}\)
Let the tension between \(m_1\) and \(m_2\) be \(T_1\), and acceleration of the system be \(a\).
Using Newton's second law: \(m_1g\sin70^{\circ}-T_1 - \mu_km_1g\cos70^{\circ}=m_1a\)

For \(m_2 = 3\space kg\):
Normal force \(N_2=m_2g\cos70^{\circ}\), frictional force \(f_2=\mu_kN_2=\mu_km_2g\cos70^{\circ}\)
Let the tension between \(m_2\) and \(m_3\) be \(T_2\)
Using Newton's second law: \(T_1 - T_2-m_2g\sin70^{\circ}-\mu_km_2g\cos70^{\circ}=m_2a\)

For \(m_3 = 1\space kg\):
Normal force \(N_3=m_3g\cos50^{\circ}\), frictional force \(f_3=\mu_kN_3=\mu_km_3g\cos50^{\circ}\)
Using Newton's second law: \(T_2 - m_3g\sin50^{\circ}-\mu_km_3g\cos50^{\circ}=m_3a\)

Step2: Substitute values and solve the system of equations

Take \(g = 9.8\space m/s^{2}\), \(\mu_k=0.1\)
\(\cos70^{\circ}\approx0.342\), \(\sin70^{\circ}\approx0.940\), \(\cos50^{\circ}\approx0.643\), \(\sin50^{\circ}\approx0.766\)

For \(m_1\): \(5\times9.8\times0.940-T_1-0.1\times5\times9.8\times0.342 = 5a\)
\(46.06 - T_1- 1.6734=5a\)
\(44.3866 - T_1=5a\)

For \(m_2\): \(T_1 - T_2-3\times9.8\times0.940 - 0.1\times3\times9.8\times0.342=3a\)
\(T_1 - T_2-27.636-1.00548 = 3a\)
\(T_1 - T_2-28.64148=3a\)

For \(m_3\): \(T_2-1\times9.8\times0.766-0.1\times1\times9.8\times0.643=a\)
\(T_2 - 7.5068-0.63014=a\)
\(T_2-8.13694=a\)

Substitute \(T_2=a + 8.13694\) into the equation for \(m_2\):
\(T_1-(a + 8.13694)-28.64148=3a\)
\(T_1-4a=36.77842\)

From the equation for \(m_1\): \(T_1=44.3866 - 5a\)

Substitute \(T_1\) into \(T_1-4a=36.77842\):
\(44.3866-5a-4a=36.77842\)
\(-9a=36.77842 - 44.3866\)
\(-9a=-7.60818\)
\(a=\frac{7.60818}{9}\approx0.845\space m/s^{2}\)

Step3: Find \(T_2\)

\(T_2=a + 8.13694\approx0.845+8.13694 = 8.98194\approx9.0\space N\)

Step4: Find \(T_1\)

\(T_1=44.3866-5a\approx44.3866-5\times0.845=44.3866 - 4.225=40.1616\approx40.2\space N\)

Answer:

The tension between \(m_1\) and \(m_2\) (\(T_1\)) is approximately \(40.2\space N\) and the tension between \(m_2\) and \(m_3\) (\(T_2\)) is approximately \(9.0\space N\)