Learn newton’s second law of motion examples, tension in a rope when a man climbs on it with some acceleration and apparent weight in an elevator. Practice to find the tension in the string connecting the two blocks.
\(\vec a = \dfrac{\vec F_{net}}{m}\)
\(\vec a =\) acceleration of body
\(m\) = mass of the object in kg
\(\Sigma F = ma\)
\(\vec F = ma\)
Acceleration, \(a = \dfrac{\vec F}{m}\)
By visual inspection, acceleration is in horizontal direction.
\(\therefore \, N + F \,sin\theta = mg\)
\(\Sigma F_y=0\)
\(\Sigma F_x =F \,cos\theta\)
Acceleration, \(a= \dfrac{\Sigma F_x}{m}\)
\(a= \dfrac{Fcos\theta}{m}\)
A \(2 \,m/s^2\)
B \(\dfrac{1}{2}\,m/s^2\)
C \(3 \,m/s^2\)
D \(\dfrac{3}{4}\) \(m/s^2\)
\( F_{A \, on\, B} = F_{B \, on \,A}\) (Action - reaction pair)
\(\vec F_{A \,on \, H} = \vec F_{H \,on \, A} > \vec F_{B \,on \, A} = \vec F_{A \,on \, B}\)
A \(N = \dfrac{m_2 F}{m_1 + m_2}\)
B \(N = \dfrac{m_1 F}{m_1 + m_2}\)
C \(N = \dfrac{ F}{m_1 + m_2}\)
D \(N = \dfrac{m_1 \,m_2}{F}\)
Free body diagrams
\(\Sigma F _x = T\)
Applying Newton's second law
\(\Sigma F _x = m_1a\)
\(T = m_1a\) ........(1)
\(\Sigma F _x = F-T\)
Applying Newton's second law
\(\Sigma F _x = m_2a\)
\(F-T = m_2a\) ........(2)
\(F- m_1a = m_2a\)
\(\dfrac{F}{m_1+m_2} = a\)
Thus, tension is given by,
\(T = \dfrac{m_1F}{m_1+m_2}\)
Then using Newton's second law
\(N = mg \)
\(\Sigma F_y = N -mg\)
\(N - mg = ma\)
\(N= mg + ma\)
\(N= m(a+ g)\)
\(\Sigma F_y = T - mg\)
T – mg = ma [Apply Newton's second law]
T = ma + mg
T = m (a + g)
T > mg
\(\lambda = \dfrac{m}{\ell}\)
F = ma
\(a = \dfrac{F}{m}\)
Applying Newton's second law
\(\Sigma F = ma\)
\(F - T = (\lambda x) \,a\)
(The value of acceleration \(a = \dfrac{F}{m}\) will remain same at this distance)
\(F - T = \dfrac{m}{\ell} × \dfrac{F}{m} ×x\)
\(F - \dfrac{F\,x}{\ell} = T \)
\(F \left( 1-\dfrac{x}{\ell}\right) = T \)
\(F \left( \dfrac{\ell - x}{\ell}\right) = T \)
Applying Newton's second law
T – mg = ma
T = mg + ma
T = m (g + a)