An object with a mass of #5 kg# is hanging from an axle with a radius of #30 m#. If the wheel attached to the axle has a radius of #9 m#, how much force must be applied to the wheel to keep the object from falling?
The force is
The load Radius of axle Radius of wheel Taking moments about the center of the axle The force is
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The force required to keep the object from falling is 14.44 N.
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To calculate the force required to keep the object from falling, we can use the principle of torque equilibrium. The force required can be calculated using the formula:
[ F = \frac{m \cdot g \cdot R}{r} ]
Where:
- ( F ) is the force required to keep the object from falling,
- ( m ) is the mass of the object (5 kg),
- ( g ) is the acceleration due to gravity (approximately 9.8 m/s²),
- ( R ) is the radius of the axle (30 m), and
- ( r ) is the radius of the wheel (9 m).
Plugging in the values:
[ F = \frac{5 , \text{kg} \times 9.8 , \text{m/s}^2 \times 30 , \text{m}}{9 , \text{m}} ]
[ F = \frac{1470 , \text{N}}{9} ]
[ F \approx 163.33 , \text{N} ]
So, approximately 163.33 Newtons of force must be applied to the wheel to keep the object from falling.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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