theo định luật II niu tơn
m2 : \(\overrightarrow{F_{ms2}}+\overrightarrow{T_2}+\overrightarrow{P_2}+\overrightarrow{Q_2}=m_2.\overrightarrow{a}_2\)
m1 : \(\overrightarrow{F_{ms1}}+\overrightarrow{F_{ms2}}+\overrightarrow{N_2}+\overrightarrow{F}+\overrightarrow{Q_1}+\overrightarrow{T_1}=m_1.\overrightarrow{a_1}\)
chiếu lên trục đã chọn
T2-Fms2=m2.a2 ; Q2=P2=m2.g
\(\Rightarrow T_2-\mu_2.m_2.g=m_2.a_2\)
F-Fms1-Fms2-T1=m1.a1 ; Q1=N2+P2=g.(m2+m1)
\(F-\mu_1.\left(m_1+m_2\right).g-\mu_2.m_2.g-T_1=m_1.a_1\)
ta có T1=T2=T; a1=a2=a, m1=m2=m; \(\mu_1=\mu_2=\mu\)
\(\Rightarrow T-\mu.m.g=m.a\)\(\Rightarrow T=m.\left(\mu.g+a\right)\) (1)
\(F-\mu.g.\left(m+m\right)-\mu.m.g-T=m.a\)
\(\Leftrightarrow F-3.\mu.m.g-T=m.a\) (2)
từ (1),(2)
F=m.a+4\(\mu\).m.g
\(\Rightarrow F=m.\left(a+4\mu g\right)\)\(\Rightarrow m=\dfrac{F}{a+4\mu g}\)
T=\(\dfrac{F}{a+4\mu.g}.\left(\mu g+a\right)\)
\(\Rightarrow T=\)\(\dfrac{120+60a}{a+8}\)
:( còn mỗi ẩn a mà sử dụng hết dữ kiện rồi!!
