# Question Video: Determining Changes in Weight due to Vertical Acceleration in a Uniform Gravitational Field

A man with a mass of 75.0 kg measures his weight by standing on a bathroom scale. The man and the scale are inside an elevator that can move at different rates. What is the measurement of the man’s weight according to the scale if the elevator is accelerating vertically upward at a rate of 1.20 m/s²? What is the measurement of the man’s weight according to the scale if the elevator is moving vertically upward at a speed of 1.20 m/s? What is the measurement of the man’s weight according to the scale if the elevator is accelerating vertically downward at a rate of 1.20 m/s²?

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### Video Transcript

A man with a mass of 75.0 kilograms measures his weight by standing on a bathroom scale. The man and the scale are inside an elevator that can move at different rates. What is the measurement of the man’s weight according to the scale if the elevator is accelerating vertically upward at a rate of 1.20 meters per second squared? What is the measurement of the man’s weight according to the scale if the elevator is moving vertically upward at a speed of 1.20 meters per second? What is the measurement of the man’s weight according to the scale if the elevator is accelerating vertically downward at a rate of 1.20 meters per second squared?

We’re told the man’s mass in this example is 75.0 kilograms. We’ll call that 𝑚. In each of these three examples, we have a man standing on a scale inside an elevator that’s either moving up or downward. If we define motion up as motion in the positive direction. And if we call 𝑊 sub 𝑠 the weight of the man as recorded by the scale in the accelerating or constantly moving elevator. Then we want to solve for that weight reading of the scale under three conditions. First, when the elevator is accelerating at positive 1.20 meters per second squared. Second, when it’s moving at a constant velocity of positive 1.20 meters per second. And lastly, we want to solve for the reading on the scale when the elevator’s acceleration is negative 1.20 meters per second squared.

What’s interesting about this scenario, looking back over to our diagram, is that if we were to draw the forces acting on the man, in all three cases, there are only two. There’s the gravitational force acting on the man, creating a weight force down on the scale. And then there’s the normal force of the scale, pushing up on the man. This normal force is equal in magnitude to the reading of the scale, 𝑊 sub 𝑠.

To get started on our first solution, we can recall Newton’s second law of motion, which tells us that the net force on an object is equal to its mass times its acceleration. In our case, that object is the man. And when the elevator is accelerating vertically upward at 1.20 meters per second squared, we can write that the normal force acting on the man minus his weight force is equal to his mass multiplied by his acceleration. Since the man is at rest with respect to the elevator, we know that his acceleration matches the elevator’s, 1.20 meters per second squared, in the positive direction.

Because we’re given the man’s mass, we also know the weight force he applies. And we know that the normal force exerted on the man is equal to the reading of the scale in this case, which is what we want to solve for. Recalling that the weight force of an object is equal to its mass times the acceleration due to gravity, we can rewrite our equation and then rearrange to solve for 𝑊 sub 𝑠.

If we treat 𝑔 as exactly 9.8 meters per second squared, then we can plug in for 𝑚, 𝑔, and 𝑎 and solve for 𝑊 sub 𝑠. When we do, we find a value of 825 newtons. That’s the weight of the man registered by the scale when the elevator accelerates upward 1.20 meters per second squared.

Next, we want to solve for the scale’s reading when the elevator is moving at a constant speed upward at 1.20 meters per second. In this case, looking back at our free body diagram of the man, we can write that the normal force acting on the man minus his weight force is equal to his mass times his acceleration, which is zero since the elevator is not accelerating. This means that the normal force is equal to the weight force. And we’ve said that the normal force is equal to the scale reading, 𝑊 sub 𝑠. So 𝑊 sub 𝑠 in this condition is equal to 𝑚 times 𝑔, or 75.0 kilograms times 9.8 meters per second squared. This is simply the weight of the man, 735 newtons.

Finally, we want to solve for the scale’s reading when the elevator is accelerating downward at a rate of 1.20 meters per second squared. Starting with the equation 𝑁, the normal force, minus 𝑊, the weight force, is equal to 𝑚 times 𝑎, we can substitute in 𝑊𝑠 for 𝑁 and rearrange so that 𝑊𝑠 is equal to 𝑚 times the quantity 𝑔 plus 𝑎. When we plug in for these values, this time we have a value of negative 1.20 for 𝑎. And when we calculate 𝑊 sub 𝑠, we find it’s equal to 645 newtons. That’s the scale’s reading when the elevator accelerates downward. We see from these three results how the scale’s reading depends on the acceleration of the elevator.