Question

Block B in rests on a surface for which the static and kinetic coefficients of friction are 0.59 and 0.40, respectively. The ropes are massless.What is the maximum mass of block A for which the system remains in static equilibrium?

Answer 1

$m_A = 11.8\:\text{kg}$

Explanation:

Let $T_A\:\text{and}\:T_B$ be the tensions on ropes on blocks A and B, respectively, $T_C$ be the tension at an angle and $f_s$ be the frictional force. Adopt the usual sign convention (right and up are positive and left and down are negative). To achieve static equilibrium, we demand that the net forces acting on the objects to be zero. Now let's apply Newton's 2nd law (NSL) to the objects.

Block A:

$y:\:\:\:T_A - m_Ag = 0$ (1)

Block B:

$x:\:\:\:T_B - f_s = T_B - \mu_sN = 0$ (2)

$y:\:\:\:N - m_Bg = 0$ (3)

Intersection of the ropes:

$x:\:\:\:T_C\cos{45} - T_B = 0 \Rightarrow T_B = T_C\cos{45}$ (4)

$y:\:\:\:T_C\sin{45} - T_A = 0 \Rightarrow T_A = T_C\sin{45}$ (5)

Using Eq(4) and Eq(3) on Eq(2) and Eq(5) on Eq(1), we get

$T_C\cos{45} = \mu m_Bg$ (6)

$T_C\sin{45} = m_Ag$ (7)

Dividing Eq(7) by Eq(6), we get

$\tan{45} = \dfrac{m_A}{\mu m_B} = 1$

Solving for $m_A$, we find that the mass needed to keep the blocks steady and motionless is

$m_A = \mu m_B = (0.59)(20\:\text{kg})$

$\:\:\:\:\:\:\:= 11.8\:\text{kg}$