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A block of mass m is pushed by applying a force F at an angle θ with the horizontal surface. The normal force on the block is given as –
- a)F = mg – F sin θ
- b)F = mg + F sin θ
- c)F = F sin θ
- d)F = mg
Correct answer is option 'A'. Can you explain this answer?
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A block of mass m is pushed by applying a force F at an angle θ w...
Both of them are vector quantities. And both of them can be easily simplified. If taken in the vector form then the task is even easier. Thus it is not necessary for the force or the couple to be vector only, even if the magnitude is taken, the simplification is done in the 2D.
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A block of mass m is pushed by applying a force F at an angle θ w...
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A block of mass m is pushed by applying a force F at an angle θ w...
Introduction:
This question is related to Newton's laws of motion and the concept of forces acting on an object. We need to determine the correct expression for the force applied on a block of mass m when it is pushed at an angle θ with the horizontal surface.
Explanation:
When a block is placed on a horizontal surface, it experiences various forces. The two main forces acting on the block are the gravitational force (mg) pulling it downwards and the normal force (N) exerted by the surface in the upward direction.
Force applied on the block:
When a force F is applied on the block at an angle θ with the horizontal surface, it can be resolved into two components:
- The vertical component (F sin θ) acts perpendicular to the surface.
- The horizontal component (F cos θ) acts parallel to the surface.
Normal force:
The normal force (N) exerted by the surface balances the vertical component of the applied force (F sin θ) to keep the block in equilibrium. Therefore, the normal force N is equal to F sin θ.
Correct expression for the applied force:
Since the normal force is equal to the vertical component of the applied force (N = F sin θ), we can rearrange this equation to find the expression for the applied force:
F = N / sin θ
Substituting the value of N:
From the earlier explanation, we know that N = F sin θ. Substituting this value into the expression for the applied force, we get:
F = (F sin θ) / sin θ
Simplifying the expression:
The sin θ terms cancel out, leaving us with:
F = F
Conclusion:
Therefore, the correct expression for the force applied on the block when it is pushed at an angle θ with the horizontal surface is F = mg sin θ, as given in option A.
This question is related to Newton's laws of motion and the concept of forces acting on an object. We need to determine the correct expression for the force applied on a block of mass m when it is pushed at an angle θ with the horizontal surface.
Explanation:
When a block is placed on a horizontal surface, it experiences various forces. The two main forces acting on the block are the gravitational force (mg) pulling it downwards and the normal force (N) exerted by the surface in the upward direction.
Force applied on the block:
When a force F is applied on the block at an angle θ with the horizontal surface, it can be resolved into two components:
- The vertical component (F sin θ) acts perpendicular to the surface.
- The horizontal component (F cos θ) acts parallel to the surface.
Normal force:
The normal force (N) exerted by the surface balances the vertical component of the applied force (F sin θ) to keep the block in equilibrium. Therefore, the normal force N is equal to F sin θ.
Correct expression for the applied force:
Since the normal force is equal to the vertical component of the applied force (N = F sin θ), we can rearrange this equation to find the expression for the applied force:
F = N / sin θ
Substituting the value of N:
From the earlier explanation, we know that N = F sin θ. Substituting this value into the expression for the applied force, we get:
F = (F sin θ) / sin θ
Simplifying the expression:
The sin θ terms cancel out, leaving us with:
F = F
Conclusion:
Therefore, the correct expression for the force applied on the block when it is pushed at an angle θ with the horizontal surface is F = mg sin θ, as given in option A.
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A block of mass m is pushed by applying a force F at an angle θ with the horizontal surface. The normal force on the block is given as –a)F = mg – F sin θb)F= mg + F sin θc)F = F sinθd)F = mgCorrect answer is option 'A'. Can you explain this answer?
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