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A 5 kg uniform serving table is supported on each side by two pairs of identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg and is in equilibrium when θ = 45°, determine the stiffness k of each spring. The springs are upstretched when θ = 90°. Neglect the mass of the links.
- a)138 N/m
- b)194 N/m
- c)166 N/m
- d)88 N/m
Correct answer is option 'C'. Can you explain this answer?
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A 5 kg uniform serving table is supported on each side by two pairs of...
Virtual Work -
- Virtual Work is the work done by a real force acting through a virtual displacement.
- A virtual displacement is any displacement consistent with the constraints of the system/structure.
- Principle of Virtual Work states that "The principle of Virtual Work states that "Total work of external forces for small virtual displacement will be zero if the rigid body is in equilibrium."
Calculation:
Given:
When θ undergoes a positive virtual displacement of δθ, the configuration given below is formed. In this figure only the spring force Fsp, the weight of table Wt and the weight of bowl Wb do work when the virtual displacement takes place.
The magnitude of Fsp can be given using the spring force formula,
The position of the point of application of Wb, Wt and Fsp are denoted by yGp, yGt and xc respectively.
From the Concept of Virtual Work,
Since Wb, Wt, and Fsp act towards the negative sense of their corresponding virtual displacement, their work is negative.
Put the value of
and Fsp = 0.25kcosθ N and all other values in equation (iv), then we have
−4.905(0.25cosθδθ) − 24.525(0.25cosθδθ) − 0.25kcosθ(−0.25sinθδθ) = 0
δθ(−7.3575cosθ + 0.0625ksinθcosθ) = 0
Since δθ ≠ 0 then
When θ = 45°, then
k = 166.48 N/m ≃ 166 N/m
Given:
When θ undergoes a positive virtual displacement of δθ, the configuration given below is formed. In this figure only the spring force Fsp, the weight of table Wt and the weight of bowl Wb do work when the virtual displacement takes place.
The magnitude of Fsp can be given using the spring force formula,
The position of the point of application of Wb, Wt and Fsp are denoted by yGp, yGt and xc respectively.
From the Concept of Virtual Work,
Since Wb, Wt, and Fsp act towards the negative sense of their corresponding virtual displacement, their work is negative.
Put the value of
and Fsp = 0.25kcosθ N and all other values in equation (iv), then we have−4.905(0.25cosθδθ) − 24.525(0.25cosθδθ) − 0.25kcosθ(−0.25sinθδθ) = 0
δθ(−7.3575cosθ + 0.0625ksinθcosθ) = 0
Since δθ ≠ 0 then
When θ = 45°, then
k = 166.48 N/m ≃ 166 N/m
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A 5 kg uniform serving table is supported on each side by two pairs of identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg and is in equilibrium when θ = 45°, determine the stiffness k of each spring. The springs are upstretched when θ = 90°. Neglect the mass of the links.a)138 N/mb)194 N/mc)166 N/md)88 N/mCorrect answer is option 'C'. Can you explain this answer?
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A 5 kg uniform serving table is supported on each side by two pairs of identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg and is in equilibrium when θ = 45°, determine the stiffness k of each spring. The springs are upstretched when θ = 90°. Neglect the mass of the links.a)138 N/mb)194 N/mc)166 N/md)88 N/mCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 5 kg uniform serving table is supported on each side by two pairs of identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg and is in equilibrium when θ = 45°, determine the stiffness k of each spring. The springs are upstretched when θ = 90°. Neglect the mass of the links.a)138 N/mb)194 N/mc)166 N/md)88 N/mCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 5 kg uniform serving table is supported on each side by two pairs of identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg and is in equilibrium when θ = 45°, determine the stiffness k of each spring. The springs are upstretched when θ = 90°. Neglect the mass of the links.a)138 N/mb)194 N/mc)166 N/md)88 N/mCorrect answer is option 'C'. Can you explain this answer?.
A 5 kg uniform serving table is supported on each side by two pairs of identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg and is in equilibrium when θ = 45°, determine the stiffness k of each spring. The springs are upstretched when θ = 90°. Neglect the mass of the links.a)138 N/mb)194 N/mc)166 N/md)88 N/mCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 5 kg uniform serving table is supported on each side by two pairs of identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg and is in equilibrium when θ = 45°, determine the stiffness k of each spring. The springs are upstretched when θ = 90°. Neglect the mass of the links.a)138 N/mb)194 N/mc)166 N/md)88 N/mCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 5 kg uniform serving table is supported on each side by two pairs of identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg and is in equilibrium when θ = 45°, determine the stiffness k of each spring. The springs are upstretched when θ = 90°. Neglect the mass of the links.a)138 N/mb)194 N/mc)166 N/md)88 N/mCorrect answer is option 'C'. Can you explain this answer?.
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