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A cord connects two bodies of weights 400N and 800N. The two bodies are placed on an inclined plane. The coefficient of friction for the weight of 400N is 0.15 and that for 800N is0.4. determine the inclination of the plane to the horizontal and the tension in the cord when the motion is about to take place down the inclined plane. the body weighing 400N is below the body weighing 800N?
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A cord connects two bodies of weights 400N and 800N. The two bodies ar...
**Solution:**
To determine the inclination of the plane and the tension in the cord when the motion is about to take place down the inclined plane, we need to consider the forces acting on the bodies and the conditions for equilibrium.
**Forces Acting on the Bodies:**
1. Body with weight 400N:
- Weight (W1) = 400N
- Frictional force (F1) = coefficient of friction (µ1) * normal force (N1)
- Normal force (N1) = weight (W1) * cos(θ)
- Tension in the cord (T) = force required to move the body down the inclined plane
2. Body with weight 800N:
- Weight (W2) = 800N
- Frictional force (F2) = coefficient of friction (µ2) * normal force (N2)
- Normal force (N2) = weight (W2) * cos(θ)
- Tension in the cord (T) = force required to move the body down the inclined plane
**Conditions for Equilibrium:**
For the motion to be about to take place down the inclined plane, the following conditions must be satisfied:
1. The net force acting on each body must be zero.
2. The frictional force acting on each body must be equal to or less than the product of the coefficient of friction and the normal force.
**Inclination of the Plane:**
To determine the inclination of the plane, we can equate the net force acting on each body to zero.
For the body with weight 400N:
- Net force = T - F1 - W1 * sin(θ) = 0
For the body with weight 800N:
- Net force = T - F2 - W2 * sin(θ) = 0
From the above equations, we can solve for the value of θ, the inclination of the plane.
**Tension in the Cord:**
To determine the tension in the cord, we can equate the frictional force acting on each body to the product of the coefficient of friction and the normal force.
For the body with weight 400N:
- F1 = µ1 * N1
For the body with weight 800N:
- F2 = µ2 * N2
From the above equations, we can solve for the value of T, the tension in the cord.
**Explanation of the Body Arrangement:**
The body weighing 400N is below the body weighing 800N because when the motion is about to take place down the inclined plane, the force of gravity acting on the body with weight 800N is larger than the force of gravity acting on the body with weight 400N. Therefore, to maintain equilibrium, the body with weight 800N needs to be placed above the body with weight 400N.
To determine the inclination of the plane and the tension in the cord when the motion is about to take place down the inclined plane, we need to consider the forces acting on the bodies and the conditions for equilibrium.
**Forces Acting on the Bodies:**
1. Body with weight 400N:
- Weight (W1) = 400N
- Frictional force (F1) = coefficient of friction (µ1) * normal force (N1)
- Normal force (N1) = weight (W1) * cos(θ)
- Tension in the cord (T) = force required to move the body down the inclined plane
2. Body with weight 800N:
- Weight (W2) = 800N
- Frictional force (F2) = coefficient of friction (µ2) * normal force (N2)
- Normal force (N2) = weight (W2) * cos(θ)
- Tension in the cord (T) = force required to move the body down the inclined plane
**Conditions for Equilibrium:**
For the motion to be about to take place down the inclined plane, the following conditions must be satisfied:
1. The net force acting on each body must be zero.
2. The frictional force acting on each body must be equal to or less than the product of the coefficient of friction and the normal force.
**Inclination of the Plane:**
To determine the inclination of the plane, we can equate the net force acting on each body to zero.
For the body with weight 400N:
- Net force = T - F1 - W1 * sin(θ) = 0
For the body with weight 800N:
- Net force = T - F2 - W2 * sin(θ) = 0
From the above equations, we can solve for the value of θ, the inclination of the plane.
**Tension in the Cord:**
To determine the tension in the cord, we can equate the frictional force acting on each body to the product of the coefficient of friction and the normal force.
For the body with weight 400N:
- F1 = µ1 * N1
For the body with weight 800N:
- F2 = µ2 * N2
From the above equations, we can solve for the value of T, the tension in the cord.
**Explanation of the Body Arrangement:**
The body weighing 400N is below the body weighing 800N because when the motion is about to take place down the inclined plane, the force of gravity acting on the body with weight 800N is larger than the force of gravity acting on the body with weight 400N. Therefore, to maintain equilibrium, the body with weight 800N needs to be placed above the body with weight 400N.
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A cord connects two bodies of weights 400N and 800N. The two bodies are placed on an inclined plane. The coefficient of friction for the weight of 400N is 0.15 and that for 800N is0.4. determine the inclination of the plane to the horizontal and the tension in the cord when the motion is about to take place down the inclined plane. the body weighing 400N is below the body weighing 800N?
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A cord connects two bodies of weights 400N and 800N. The two bodies are placed on an inclined plane. The coefficient of friction for the weight of 400N is 0.15 and that for 800N is0.4. determine the inclination of the plane to the horizontal and the tension in the cord when the motion is about to take place down the inclined plane. the body weighing 400N is below the body weighing 800N? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A cord connects two bodies of weights 400N and 800N. The two bodies are placed on an inclined plane. The coefficient of friction for the weight of 400N is 0.15 and that for 800N is0.4. determine the inclination of the plane to the horizontal and the tension in the cord when the motion is about to take place down the inclined plane. the body weighing 400N is below the body weighing 800N? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cord connects two bodies of weights 400N and 800N. The two bodies are placed on an inclined plane. The coefficient of friction for the weight of 400N is 0.15 and that for 800N is0.4. determine the inclination of the plane to the horizontal and the tension in the cord when the motion is about to take place down the inclined plane. the body weighing 400N is below the body weighing 800N?.
A cord connects two bodies of weights 400N and 800N. The two bodies are placed on an inclined plane. The coefficient of friction for the weight of 400N is 0.15 and that for 800N is0.4. determine the inclination of the plane to the horizontal and the tension in the cord when the motion is about to take place down the inclined plane. the body weighing 400N is below the body weighing 800N? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A cord connects two bodies of weights 400N and 800N. The two bodies are placed on an inclined plane. The coefficient of friction for the weight of 400N is 0.15 and that for 800N is0.4. determine the inclination of the plane to the horizontal and the tension in the cord when the motion is about to take place down the inclined plane. the body weighing 400N is below the body weighing 800N? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cord connects two bodies of weights 400N and 800N. The two bodies are placed on an inclined plane. The coefficient of friction for the weight of 400N is 0.15 and that for 800N is0.4. determine the inclination of the plane to the horizontal and the tension in the cord when the motion is about to take place down the inclined plane. the body weighing 400N is below the body weighing 800N?.
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A cord connects two bodies of weights 400N and 800N. The two bodies are placed on an inclined plane. The coefficient of friction for the weight of 400N is 0.15 and that for 800N is0.4. determine the inclination of the plane to the horizontal and the tension in the cord when the motion is about to take place down the inclined plane. the body weighing 400N is below the body weighing 800N?, a detailed solution for A cord connects two bodies of weights 400N and 800N. The two bodies are placed on an inclined plane. The coefficient of friction for the weight of 400N is 0.15 and that for 800N is0.4. determine the inclination of the plane to the horizontal and the tension in the cord when the motion is about to take place down the inclined plane. the body weighing 400N is below the body weighing 800N? has been provided alongside types of A cord connects two bodies of weights 400N and 800N. The two bodies are placed on an inclined plane. The coefficient of friction for the weight of 400N is 0.15 and that for 800N is0.4. determine the inclination of the plane to the horizontal and the tension in the cord when the motion is about to take place down the inclined plane. the body weighing 400N is below the body weighing 800N? theory, EduRev gives you an
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