Weight in kg!

John is standing on a weighing machine installed in a lift. It shows a reading of 50 kg 50\text{ kg} when the lift is at rest.

What will happen to the reading of the machine if the lift starts accelerating downward?

It will decrease It will increase It will remain the same

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4 solutions

Pranshu Gaba
Dec 15, 2017

Relevant wiki: Newton's Second Law

Although the weighing machine displays the reading in kilograms, it does not measure the mass. but the force by which it is being pressed.

When the lift starts accelerating downward, it is equivalent to the gravity in the lift becoming weaker. As a result, John does not press against the machine as hard, and the reading of the machine decreases.

This can also be seen using Newton's second law of motion. Let the mass of John be m m . His weight is m g mg . When he is rest in the lift, the machine exerts normal force N N upwards on John. The machine shows this reading N N . Since he is not accelerating, using Newton's second law of motion, m g N = 0 mg - N = 0 .

When John is accelerating downwards with acceleration a a , the equation of motion is m g N = m a mg - N = ma . The normal force now is N = m g m a N = mg - ma , which is less than m g mg . We see that the reading of the machine reduced.

Hana Wehbi
Nov 25, 2017

Relevant wiki: Newton's Laws of Motion

In this case, the elevator and the person are initially moving upward at a constant speed and slowing down to rest at a higher floor. The acceleration of the elevator is downward (opposite to the upward motion, which causes a reduction of the velocity). The inertia of the person would prefer to keep moving upward at a constant speed, so the elevator floor and scale effectively drop out a little bit from underneath the person as the elevator slows down.

The person doesn't float upward, because again the elevator and the person move together, but the contact force between the person and the scale is reduced. The scale therefore has to push upward with less force on the person to support the person's weight. Therefore the Normal Force is smaller, so the reading on the scale is a number that is LESS than the true weight.

An other way is to think of acceleration and gravity as providing the same effect on the observer. A lift accelerating upwards at (/9.8ms^-1/) would same as the Earth's gravitational pull to the observer As the lift is accelerating downwards the PULL of the Earth and the lesser PUSH of the lift almost get cancelled out giving a lesser count of Net force

Kartik Jay - 3 years, 6 months ago

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Nice alternative solution.

Hana Wehbi - 3 years, 6 months ago
Jacky Blond
Nov 27, 2017

人一开始静止在称上,所以人收到的重力与称给人的支持力一样大小,当电梯开始运动时,有向下的加速度a,由牛顿第二定律可知a=F合/质量M,所以此时有合外力,又因为加速度方向向下,可知是人所受的重力大于称给人的支持力,所以称的视数要小于重力大小 ,所以视数变小。৫(”ړ৫) my English isn't good. so i use Chinese to show my solution

English Translation

When the elevator begins to move, there is a downward acceleration a. The Newton's second law shows that a = F / mass M, So at this time there are co-forces, but also because the acceleration direction down, we can see that the human body is greater than the force of gravity to support people, so that the number of views is less than the size of gravity, so the visual number becomes smaller.

English translation:

When the elevator begins to move, there is a downward acceleration a. The Newton's second law shows that a = F / mass M , So at this time there are co-forces, but also because the acceleration direction down, we can see that the human body is greater than the force of gravity to support people, so that the number of views is less than the size of gravity, so the visual number becomes smaller. 5 ("ړ 5)

Munem Shahriar - 3 years, 6 months ago
Gregory Lewis
Nov 28, 2017

Although kg is a measurement of mass, not weight, it was stated that it was a weighing machine displaying its results in kg (assuming Earth gravity, presumably) and not a true measure of kg. Therefore, we can assume it is the weight and not the mass that is the basis of the displayed value.

The acceleration of the elevator can be treated like a decrease in gravity, decreasing the weight and therefore the value displayed, which will no longer be technically correct.

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