An object with a mass of #8 kg# is hanging from an axle with a radius of #15 m#. If the wheel attached to the axle has a radius of #90 m#, how much force must be applied to the wheel to keep the object from falling?

Answer 1

The force is #=13.1N#

The load is #L=8gN#

The radius of the axle is #r=15m#

The radius of the wheel is #R=90m#

The effort is #=FN#

The acceleration due to gravity is #g=9.8ms^-2#

Taking moments about the center of the axle

#F*90=8g*15#

#F=8g*15/90=13.1N#

The force is #F=13.1N#

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Answer 2

To calculate the force required to keep the object from falling, we need to consider the torque exerted by the object and the wheel.

The torque exerted by the object hanging from the axle is given by:

[ \text{Torque}\text{object} = \text{Force}\text{object} \times \text{Radius}_\text{axle} ]

The torque exerted by the wheel is given by:

[ \text{Torque}\text{wheel} = \text{Force}\text{wheel} \times \text{Radius}_\text{wheel} ]

Since the object is not falling, the torques must be equal:

[ \text{Torque}\text{object} = \text{Torque}\text{wheel} ]

[ \text{Force}\text{object} \times \text{Radius}\text{axle} = \text{Force}\text{wheel} \times \text{Radius}\text{wheel} ]

Solving for the force required on the wheel (( \text{Force}_\text{wheel} )):

[ \text{Force}\text{wheel} = \frac{\text{Force}\text{object} \times \text{Radius}\text{axle}}{\text{Radius}\text{wheel}} ]

Given:

  • Mass of the object (( m )) = 8 kg
  • Radius of the axle (( r_\text{axle} )) = 15 m
  • Radius of the wheel (( r_\text{wheel} )) = 90 m
  • Acceleration due to gravity (( g )) = 9.8 m/s²

First, calculate the force exerted by the object using Newton's second law:

[ \text{Force}_\text{object} = m \times g ]

[ \text{Force}_\text{object} = 8 \times 9.8 ]

[ \text{Force}_\text{object} = 78.4 , \text{N} ]

Now, substitute the given values into the equation for the force required on the wheel:

[ \text{Force}_\text{wheel} = \frac{78.4 \times 15}{90} ]

[ \text{Force}_\text{wheel} = \frac{1176}{90} ]

[ \text{Force}_\text{wheel} \approx 13.067 , \text{N} ]

So, approximately 13.067 Newtons of force must be applied to the wheel to keep the object from falling.

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Answer from HIX Tutor

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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