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500 kg medical equipment trolley is pushed up an incline plane to a vertical height of 2.5m. the length of the inclined plane is 10m an effort of 1800N is required to move the trolley up the plane. a) Calculate the load due to the weight of the trolley. b) Mechanical advantage.

Physics

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Answers

2 parts
a

4905 N\text{4905 N}

b

2.725\text{2.725}

Step-by-step solution

6 steps

Part a· 3 steps

Calculate the load due to the weight of the trolley.

The load due to the weight of the trolley is its weight.

  1. State the formula for weight.

    W=m×gW = m \times g

  2. Substitute the given mass (m=500 kgm = 500 \, \text{kg}) and the acceleration due to gravity (g=9.81 m/s2g = 9.81 \, \text{m/s}^2).

    W=500 kg×9.81 m/s2W = 500 \, \text{kg} \times 9.81 \, \text{m/s}^2

  3. Calculate the weight.

    W=4905 NW = 4905 \, \text{N}

    The student's calculation of 12262.5 was for mghmgh, which represents the work done against gravity, not the load (weight) itself.
    The load is 4905 N\boxed{\text{4905 N}}.

Part b

2.725\text{2.725}

Mechanical advantage.

The mechanical advantage (specifically, the Actual Mechanical Advantage, AMA) is the ratio of the load to the effort.

  1. State the formula for Actual Mechanical Advantage (AMA).

    AMA=LoadEffortAMA = \frac{\text{Load}}{\text{Effort}}

  2. Substitute the load from part a) (4905 N4905 \, \text{N}) and the given effort force (1800 N1800 \, \text{N}).

    AMA=4905 N1800 NAMA = \frac{4905 \, \text{N}}{1800 \, \text{N}}

  3. Calculate the mechanical advantage.

    AMA=2.725AMA = 2.725

    The mechanical advantage is 2.725\boxed{\text{2.725}}.

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