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a block of mass 10kg is placed on an inclined plane when the angle of inclination is 30 degree the block just begins to slide down the plane the force of static friction?
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?a block of mass 10kg is placed on an inclined plane when the angle of...
The force of static friction can be determined by analyzing the forces acting on the block and understanding the conditions for it to start sliding down the inclined plane. In this scenario, a block of mass 10 kg is placed on an inclined plane at an angle of inclination of 30 degrees.
1. Forces acting on the block:
- Weight (W): This is the force exerted by gravity on the block and can be calculated as the product of mass (m) and acceleration due to gravity (g). In this case, W = 10 kg * 9.8 m/s^2 = 98 N.
- Normal force (N): This is the perpendicular force exerted by the inclined plane on the block. It is equal in magnitude and opposite in direction to the component of the weight acting perpendicular to the plane. In this case, N = W * cos(theta) = 98 N * cos(30 degrees) = 84.85 N.
- Force of static friction (f): This is the force exerted by the inclined plane on the block parallel to the surface, in the opposite direction of the impending motion. It prevents the block from sliding down the plane.
2. Conditions for impending motion:
For the block to start sliding down the inclined plane, the force of static friction must reach its maximum value. The maximum force of static friction (fs_max) can be calculated using the equation fs_max = μs * N, where μs is the coefficient of static friction.
3. Calculation of the force of static friction:
To calculate the force of static friction, we need to know the coefficient of static friction between the block and the inclined plane. Assuming the coefficient of static friction (μs) is known, we can substitute the values into the equation fs_max = μs * N.
4. Conclusion:
Without knowing the specific value of the coefficient of static friction, we cannot determine the exact force of static friction acting on the block. However, by using the calculated values for weight, normal force, and the equation for the maximum force of static friction, we can determine the maximum force that static friction can exert to prevent the block from sliding down the inclined plane.
1. Forces acting on the block:
- Weight (W): This is the force exerted by gravity on the block and can be calculated as the product of mass (m) and acceleration due to gravity (g). In this case, W = 10 kg * 9.8 m/s^2 = 98 N.
- Normal force (N): This is the perpendicular force exerted by the inclined plane on the block. It is equal in magnitude and opposite in direction to the component of the weight acting perpendicular to the plane. In this case, N = W * cos(theta) = 98 N * cos(30 degrees) = 84.85 N.
- Force of static friction (f): This is the force exerted by the inclined plane on the block parallel to the surface, in the opposite direction of the impending motion. It prevents the block from sliding down the plane.
2. Conditions for impending motion:
For the block to start sliding down the inclined plane, the force of static friction must reach its maximum value. The maximum force of static friction (fs_max) can be calculated using the equation fs_max = μs * N, where μs is the coefficient of static friction.
3. Calculation of the force of static friction:
To calculate the force of static friction, we need to know the coefficient of static friction between the block and the inclined plane. Assuming the coefficient of static friction (μs) is known, we can substitute the values into the equation fs_max = μs * N.
4. Conclusion:
Without knowing the specific value of the coefficient of static friction, we cannot determine the exact force of static friction acting on the block. However, by using the calculated values for weight, normal force, and the equation for the maximum force of static friction, we can determine the maximum force that static friction can exert to prevent the block from sliding down the inclined plane.
Community Answer
?a block of mass 10kg is placed on an inclined plane when the angle of...
50× sqrt of 3 N
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?a block of mass 10kg is placed on an inclined plane when the angle of inclination is 30 degree the block just begins to slide down the plane the force of static friction?
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?a block of mass 10kg is placed on an inclined plane when the angle of inclination is 30 degree the block just begins to slide down the plane the force of static friction? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about ?a block of mass 10kg is placed on an inclined plane when the angle of inclination is 30 degree the block just begins to slide down the plane the force of static friction? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ?a block of mass 10kg is placed on an inclined plane when the angle of inclination is 30 degree the block just begins to slide down the plane the force of static friction?.
?a block of mass 10kg is placed on an inclined plane when the angle of inclination is 30 degree the block just begins to slide down the plane the force of static friction? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about ?a block of mass 10kg is placed on an inclined plane when the angle of inclination is 30 degree the block just begins to slide down the plane the force of static friction? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ?a block of mass 10kg is placed on an inclined plane when the angle of inclination is 30 degree the block just begins to slide down the plane the force of static friction?.
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