Question

In ansamblul din figura se cunosc m1 = 2 KG; m2 = 3 KG si m3 = 10 KG. Aproximand g = 10 N/KG ( m/s la a 2-a ), sa se determine acceleratia corpurilor si tensiunea in fir.

Answer 1

$fie->M=m_1+m_2=2+3=5 kg\\pt.M:\vec{T}+\vec{G}=M\vec{a}\\T-Mg=Ma(1)\\pt.m_3:\vec{T}+\vec{G_3}=m_3\vec{a}\\m_3g-T=m_3a(2)\\(1)+(2)=>m_3g-Mg=a(M+m_3)<=>g(m_3-M)=a(M+m_3)=>a=\frac{g(m_3-M)}{M+m_3}= \frac{10(10-5)}{5+10}=\frac{50}{15}= 3,33 m/s^2\\(1)=>T=M(a+g)=5(10+3,33)=13,33*5=16,66 N\\pt.m_2:\vec{G_2}+\vec{T'}=m_2a\\m_2g-T'=m_2a=>T'=m_2(g-a)=3(10-3,33)=3*6,66=20N$