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Ch 05: Applying Newton's Laws
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 5, Problem 7

Find the tension in each cord in Fig. E5.75.7 if the weight of the suspended object is ww.
Diagram showing a suspended object with weight 'w' and two cords at angles of 30° and 45°, labeled A, B, and C.
Diagram showing a suspended object with weight w, supported by two cords at angles of 60° and 45°.

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Step 1: Identify the forces acting on the system. The object of weight w is suspended by three cords: cord A, cord B, and cord C. Cord C is vertical, while cords A and B are at angles of 20° and 35° respectively with the horizontal.
Step 2: Apply equilibrium conditions. Since the object is stationary, the forces in both the horizontal and vertical directions must sum to zero. This gives two equations: ΣFx = 0 (horizontal forces) and ΣFy = 0 (vertical forces).
Step 3: Write the horizontal force equation. The horizontal components of the tensions in cords A and B must cancel each other out: T_A * cos(20°) = T_B * cos(35°).
Step 4: Write the vertical force equation. The vertical components of the tensions in cords A and B must balance the weight w and the tension in cord C: T_A * sin(20°) + T_B * sin(35°) = w.
Step 5: Solve the system of equations. Use the two equations derived in steps 3 and 4 to solve for the tensions T_A and T_B in cords A and B. Cord C's tension is equal to the weight w since it directly supports the object vertically.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Tension in Cords

Tension is the force exerted along a cord or rope when it is pulled tight by forces acting from opposite ends. In this scenario, the tensions in cords A and B must balance the weight of the suspended object (w) while also considering the angles at which the cords are positioned. The tension can be calculated using the components of the forces acting in both the vertical and horizontal directions.
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Equilibrium of Forces

For an object to be in equilibrium, the sum of all forces acting on it must equal zero. This means that the vertical components of the tensions in the cords must equal the weight of the object, while the horizontal components must cancel each other out. This principle allows us to set up equations based on the angles and the weight to solve for the tensions in the cords.
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Trigonometric Functions

Trigonometric functions, such as sine and cosine, relate the angles of a triangle to the ratios of its sides. In this problem, these functions are essential for resolving the tension forces into their vertical and horizontal components. For example, the vertical component of the tension in cord A can be found using T_A * sin(20°), and similarly for cord B using T_B * sin(35°).
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Related Practice
Textbook Question

A large wrecking ball is held in place by two light steel cables (Fig. E5.65.6). If the mass m of the wrecking ball is 36203620 kg, what are the (a) tension TBT_B in the cable that makes an angle of 40°40° with the vertical and (b) the tension TAT_{A} in the horizontal cable?

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Textbook Question

A picture frame hung against a wall is suspended by two wires attached to its upper corners. If the two wires make the same angle with the vertical, what must this angle be if the tension in each wire is equal to 0.750.75 of the weight of the frame? (Ignore any friction between the wall and the picture frame.)

Textbook Question

A 75.075.0-kg wrecking ball hangs from a uniform, heavy-duty chain of mass 26.026.0 kg. What is the tension at a point three-fourths of the way up from the bottom of the chain?

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Textbook Question

A 11301130-kg car is held in place by a light cable on a very smooth (frictionless) ramp (Fig. E5.85.8). The cable makes an angle of 31.0°31.0° above the surface of the ramp, and the ramp itself rises at 25.0°25.0° above the horizontal. How hard does the surface of the ramp push on the car?

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Textbook Question

A 11301130-kg car is held in place by a light cable on a very smooth (frictionless) ramp (Fig. E5.85.8). The cable makes an angle of 31.0°31.0° above the surface of the ramp, and the ramp itself rises at 25.0°25.0° above the horizontal. Find the tension in the cable.

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Textbook Question

A 11301130-kg car is held in place by a light cable on a very smooth (frictionless) ramp (Fig. E5.85.8). The cable makes an angle of 31.0°31.0° above the surface of the ramp, and the ramp itself rises at 25.0°25.0° above the horizontal. Draw a free-body diagram for the car.

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