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

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

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

Answer 2

The maximum mass of block A for which the system remains in static equilibrium with block B, resting on a surface with static and kinetic coefficients of 0.59 and 0.40, respectively, is 11.8 kg.

We can find the maximum mass of block A by the sum of forces on blocks A and B on the x and y-directions.

For the system to be in equilibrium, the sum of the forces must be equal to zero.

In the x-direction

Block B

$\Sigma F_{B}_{x} = 0$

$T_{1} - \mu_{s}N_{B} = 0$    

$T_{1} - \mu_{s}m_{B}g = 0$    (1)

Where:

• $\mu_{s}$: is the static coefficient of friction = 0.59

• $m_{B}$: is the mass of block B = 20 kg

• g: is the acceleration due to gravity = 9.81 m/s²

• T₁: is the tension force 1 (see the picture below)

• N: is the normal force = mg

Block A

For block A there are no components of the force in the x-direction.

In the y-direction

Block B

The only force acting on block B in the y-direction is the weight which is in the normal force in equation 1.            

Block A

$\Sigma F_{A}_{y} = 0$

$T_{3} - m_{A}g = 0$  (2)

To find the tension forces, we need to see all the forces acting on the node (point C in the picture below), as follows:

In x-direction

$\Sigma F_{C}_{x} = 0$

$T_{1} - T_{2}_{x}= 0$

$T_{1} - T_{2}cos(45) = 0$

$T_{2}cos(45) = T_{1}$     (3)

In y-direction                  

$\Sigma F_{C}_{y} = 0$

$T_{2}_{y} - T_{3} = 0$

$T_{2}sin(45) - T_{3} = 0$    

$T_{2}sin(45) = T_{3}$   (4)

By dividing equations (4) and (3), we have:

$\frac{T_{2}sin(45)}{T_{2}cos(45)} = \frac{T_{3}}{T_{1}}$    

$tan(45) = \frac{T_{3}}{T_{1}}$   (5)

By entering T₃ (eq 2) and T₁ (eq 1) into equation (5), we have:

$tan(45) = \frac{m_{A}g}{\mu_{s}m_{B}g}$

Solving for $m_{A}$, we have:

$m_{A} = tan(45)*\mu_{s}*m_{B} = tan(45)*0.59*20 kg = 11.8 kg$

Therefore, the maximum mass of block A for which the system remains in static equilibrium is 11.8 kg.

Find more about static coefficient of friction here:  

brainly.com/question/13758352

I hope it helps you!