Answer:
The greatest mass, that the weight C can have such that block A does not move is approximately 23.259 kg
Explanation:
The given parameters are;
The mass of the block A= 58 kg
The coefficient of static friction between the block and the surface, [tex]\mu _s[/tex] = 0.300
The coefficient of static friction between the rope and the fixed peg, B [tex]\mu _B[/tex] = 0.310
Let T represent the tension in the rope
Therefore, when the rope is static, T = The normal reaction at the peg, B, [tex]N_B[/tex]
The angle of inclination of the rope holding the block A = arctan(3/4) ≈ 36.87°
The length of the rope = √(0.4² + 0.3²) = 0.5
∴ sin(θ) = 3/5 = 0.6
cos(θ) = 4/5 = 0.8
The vertical component of the tension in the rope = T × sin(θ) = 0.6·T
The horizontal component of the tension in the rope = T × cos(θ) = 0.8·T
The friction force = μ×(W - 0.6·T) = 0.300×(58×9.8 - 0.6·T) = 170.52 - 0.18·T
The block will start to move when we have;
The horizontal component of the tension in the rope = The friction force
∴ 0.8·T = 170.52 - 0.18·T
0.8·T + 0.18·T = 170.52
0.98·T = 170.52
T = 170.52/0.98 = 174
Therefore, the tension in the rope = T = 174 N = The normal reaction at the peg, B [tex]N_B[/tex]
The frictional force at the peg, [tex]F_B[/tex] = [tex]\mu _B[/tex] × [tex]N_B[/tex] = 0.310 × 174 N = 53.94 N
The weight of the mass, [tex]m_c[/tex], [tex]W_c[/tex] = The frictional force at the peg, [tex]F_B[/tex] + The tension in the rope
∴ The weight of the mass, [tex]m_c[/tex], [tex]W_c[/tex] = 53.94 N + 174 N = 227.94 N
Weight, W = Mass, m × The acceleration due to gravity, g, from which we have;
m = W/g
Where;
g = 9.8 m/s²
∴ [tex]m_c[/tex] = [tex]W_c[/tex]/g = 227.94 N/(9.8 m/s²) ≈ 23.259 kg.
The greatest mass, that the weight C can have such that block A does not move = [tex]m_c[/tex] ≈ 23.259 kg.
