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Consider a uniform cubical box of side 'a' on a rough floor that is to be moved by applying minimum possible force F at a point 'b' above its centre of mass (see figure). If the coefficient of friction is μ = 0.4, the maximum possible value of 100 × b/a for box not to topple before moving is ______.
Correct answer is '75'. Can you explain this answer?
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Consider a uniform cubical box of side a on a rough floor that is to b...
The figure is missing, so I am unable to provide a specific answer. However, I can still provide some general information about the problem.
To find the minimum possible force required to move the cubical box, we need to consider the forces acting on it. The weight of the box acts vertically downwards through its center of mass. The normal force from the floor acts vertically upwards and is equal in magnitude to the weight of the box.
The force of friction between the box and the floor opposes the motion and depends on the coefficient of friction. The frictional force can be calculated using the equation:
Frictional force = coefficient of friction * normal force
To minimize the force required to move the box, we want to minimize the frictional force. This can be achieved by minimizing the coefficient of friction.
If the coefficient of friction is zero, there will be no frictional force and the box will move with the minimum possible force. In this case, the force applied at point b above the center of mass should be enough to overcome the inertia of the box and start its motion.
However, if the coefficient of friction is non-zero, the minimum force required to move the box will depend on the specific values of the coefficient of friction and the weight of the box.
To find the minimum possible force required to move the cubical box, we need to consider the forces acting on it. The weight of the box acts vertically downwards through its center of mass. The normal force from the floor acts vertically upwards and is equal in magnitude to the weight of the box.
The force of friction between the box and the floor opposes the motion and depends on the coefficient of friction. The frictional force can be calculated using the equation:
Frictional force = coefficient of friction * normal force
To minimize the force required to move the box, we want to minimize the frictional force. This can be achieved by minimizing the coefficient of friction.
If the coefficient of friction is zero, there will be no frictional force and the box will move with the minimum possible force. In this case, the force applied at point b above the center of mass should be enough to overcome the inertia of the box and start its motion.
However, if the coefficient of friction is non-zero, the minimum force required to move the box will depend on the specific values of the coefficient of friction and the weight of the box.
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Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force F at a point b above its centre of mass (see figure). If the coefficient of friction is μ = 0.4, the maximum possible value of 100 × b/a for box not to topple before moving is ______.Correct answer is '75'. Can you explain this answer?
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