Answer:
[tex]F=1.159[/tex]
Explanation:
From the question we are told that:
Mass of pulley [tex]M=1kg[/tex]
Radius [tex]r=12cm[/tex]
Mass of block A [tex]M_a=2.1kg[/tex]
Mass of block B [tex]m_b=4.1kg[/tex]
Spring constant[tex]\mu= 358 J/m2[/tex]
Generally the equation for Torque is mathematically given by
Since [tex]\sumF=ma[/tex]
At mass A
[tex]T_2-f_3=2.1a[/tex]
At mass B
[tex]4.8-T_1=4.1a[/tex]
At Pulley
[tex]R(T_1-T_2)=\frac{1*1*R^2}{2}\frac{a}{R}[/tex]
[tex]R(T_1-T_2)=0.55a[/tex]
Therefore the equation for total force F
At mass A+At mass B+At Pulley
[tex](T_2-f_3+4.8-T_1+R(T_1-T_2)=2.1a+4.1a+0.55a[/tex]
[tex](T_2-f_3+4.8-T_1+R(T_1-T_2)=2.7a+4.8a+0.55a[/tex]
[tex]-f_3+4.1=6.75a[/tex]
[tex]-f_3=6.75a+4.8[/tex]
Since From above equation
[tex]M_{eff}=6.7kg[/tex]
Therefore
[tex]T=2\pi \sqrt{{\frac{M_{eff}}{k}}[/tex]
[tex]T=2\pi \sqrt{{\frac{6.75}{\mu}}[/tex]
[tex]T=0.862s[/tex]
Generally the equation for frequency is mathematically given by
[tex]F=\frac{1}{T} \\F=\frac{1}{0.862}[/tex]
[tex]F=1.159[/tex]
