A uniform disk of mass m = 2.4 kg and radius R = 25 cm can rotate about an axle through its center. Four forces are acting on it
as shown in the figure. Their magnitudes are F1 = 8.5 N, F2 = 1.5 N, F3 = 6.5 N and F4 = 6.5 N. F2 and F4 act a distance d = 3.5 cm from the center of mass. These forces are all in the plane of the disk. Write an expression of the magnitude of the torque due to the force F3
1 answer:
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Answer:
The resultant tension of the two ropes is approximately 42.4 N
The length of the line representing the resultant tension is approximately 8.48 cm
Please find included with the answer the scale drawing created with Microsoft Word
Explanation:
The given parameters are;
The tension in rope P, = 30 N
The tension in rope Q, = 30 N
The angle the rope, 'P', makes with the horizontal = 45°
The angle the rope, 'Q', makes with the horizontal = 45°
The scale factor of the scale diagram, S.F. = 5.0 N/cm
By the resolution of forces at equilibrium, we have;
The sum of the vertical forces, =
+
+ W = 0
∴ W = -( +
)
W = -(30 × sin(45°) + 30 × sin(45°)) = -42.4264068712
The weight of the heavy ball, W ≈ 42.4 N acting downwards
The sum of the horizontal forces, =
+
= 0
The length of the resultant force, W = W/(S.F.) ≈ 42.4 N/(5.0 N/cm) = 8.48 cm
The drawing of the vectors using the scale factor of 5.0 N/cm is created using Microsoft Word is included
