kazah
megoldása
1,
I. `m_1*a` = F-K
II. `m_2*a` = K
I.: a = `(F-K)/m_1`
II. `m_2*(F-K)/m_1` = K
`m_2*F-m_2*K` = `m_1*K`
`m_2*F` = `(m_1+m_2)*K`
K = `(m_2*F)/(m_1+m_2)` = `(3*8)/(3+2)` = `24/5` = 4,8 N
Ábra
2,
I. `m_1*a` = `F-K_1-mu_m_1*g`
II. `m_2*a` = `K_1-K_2-mu*m_2*g`
III. `m_3*a` = `K_2-mu*m_3*g`
I. 10a = `100-K_1-0.1*10*10` = `90-K_1`
II. 7a = `K_1-K_2-0.1*7*10` = `K_1-K_2-7`
III. 3a = `K_2-0.1*3*10` = `K_2-3`
III. `K_2` = 3a+3
I. `K_1` = 90-10a
II.
7a = 90-10a-3a-3-7 = 80-13a
20a = 80
a = 4 `m/s^2`
I. : `K_1` = `90-10*4` = 90-40 = 50 N
III: `K_2` = `3*4+3` = 15 N
Ábra
1