Answer to Question #90728 in Mechanics | Relativity for Gene Alexis E. Soriano

Question

In figure blocks of mass m1=3kg and m2=4kg are connected by a string of negligible mass and are initially held in a place. The heavier block is tilted at 30 degrees. The coefficient of kinetic friction between the block of mass 3kg and the horizontal surface is 0.3. The coefficient of kinetic friction bock mass 4kg and the inclined surface is 0.1. The pulley has a negligible mass and the friction. once they are released, the blocks move, what then is the tension in the string and accelaration?

Answer 1

Apply Second Newton's Law to mass m₁:

in the x direction

$m_1 \cdot a=T-f_1=\{ f_1=\mu _1 \cdot Fn_1 \}=T-\mu _1 \cdot Fn_1 \space (1)$

in the y direction

$0=Fn_1 -m_1 \cdot g \space (2)$

From (1) and (2):

$T = m_1 \cdot a + \mu_1 \cdot m_1 \cdot g \space (3)$

Apply Second Newton's Law to mass m₂:

in the x direction

$m_2 \cdot a \cdot cos30^ \circ=-T \cdot cos30^ \circ - f_2 \cdot cos30^ \circ + Fn_2 \cdot sin30^ \circ$ $f_2 = \mu_2 \cdot Fn_2 \space \rightarrow \space m_2 \cdot a =-T + Fn_2 \cdot (1/ \sqrt{3} - \mu_2) \space (4)$

in the y direction

$-m_2 \cdot a \cdot sin30^ \circ=T \cdot sin30^ \circ + f_2 \cdot sin30^ \circ + Fn_2 \cdot cos30^ \circ - m_2 \cdot g$ $-m_2 \cdot a =T + Fn_2 \cdot (\sqrt{3}+\mu_2) - 2 \cdot m_2 \cdot g \space (5)$

From (4) and (5):

$-m_2 \cdot a =T + (m_2 \cdot a +T) \frac {\sqrt{3}+\mu_2} {1/ \sqrt{3} - \mu_2} - 2 \cdot m_2 \cdot g$ $a=\frac g {m_1+m_2} \Big( \frac {m_2} {2} (1-\sqrt{3} \cdot \mu_2)-\mu_1 \cdot m_1 \Big)$ $a=\frac {9.81 m/s^2} {3 kg+4 kg} \Big( \frac {4 kg} {2} (1-\sqrt{3} \cdot 0.1 ) - 0.3 \cdot 3kg \Big) \cong 1.06 \frac {m} {s^2}$

From (3):

$T = m_1 \cdot a + \mu_1 \cdot m_1 \cdot g = 3kg \cdot 1.06 m/s^2 + 0.3 \cdot 3kg \cdot 9.81 m/s^2 \cong 11.997 N$

Answer:

$T \cong 12 N ,\space \space a \cong 1m/s^2$