Weighing a Salmon in an Elevator: What's the Apparent Weight?

[Snowflake9]

Homework Statement

A 5.0kg salmon is weighed by hanging it from a fish scale attached to the ceiling of an elevator. What is the apparent weight of the salmon if the elevator is (a) at rest, (b) moving upward at 2.5 m/s2, and (c) moving downward at 3.2 m/s2?

Homework Equations

Sum of F = 0
F(t) - W = 0
F(t) = W

Sum of F = Ma
F(t) - W = Ma (W=mg)

F(t)<W Sum of F = Ma
W- F(t) = Ma

The Attempt at a Solution

i have trouble thinking out of the box to get the problem started. this is my very first time ever taking physics and i have not had a math or science class in a year. can anyone help me get started?! I would appreciate any help! Thank you!

 

[kjohnson]

Part 1, you are correct. Parts 2 & 3 you have to correct formula, use F=ma. Part 2 you should get the correct answer which should be greater than the static weight. I can't tell if part 3 is a trick question since I believe that would make your "tension force" in the other direction i.e. a compression force and a rope can't carry compression...

 

[kjohnson]

brain fart..on part 3 since the acceleration is less than gravity you would still need a tension force up.

 

[Snowflake9]

thank you for the response!

 

[rwisz]

I understand that the apparent weight of an object is the force exerted on it by a supporting surface, such as a scale, in order to keep it in equilibrium. In this case, the salmon is being suspended by the fish scale attached to the ceiling of the elevator.

In order to solve this problem, we must first consider the forces acting on the salmon. The only two forces acting on the salmon are its weight, which is the force due to gravity pulling it down, and the normal force, which is the force exerted by the scale to support the salmon's weight.

When the elevator is at rest, the salmon is in equilibrium and the sum of forces acting on it is zero. This means that the normal force must be equal in magnitude to the weight of the salmon, which is given as 5.0kg. Therefore, the apparent weight of the salmon in this case is also 5.0kg.

When the elevator is moving upward at 2.5 m/s^2, the salmon is experiencing an acceleration in the same direction as its weight. This means that the normal force must be greater than the weight in order to keep the salmon in equilibrium. Using the equation F(t) - W = Ma, we can solve for the normal force, which is the apparent weight of the salmon in this case. Plugging in the given values, we get F(t) = (5.0kg)(9.8m/s^2) + (5.0kg)(2.5m/s^2) = 58.5N. Therefore, the apparent weight of the salmon when the elevator is moving upward is 58.5N.

Similarly, when the elevator is moving downward at 3.2 m/s^2, the salmon is experiencing an acceleration in the opposite direction of its weight. This means that the normal force must be less than the weight in order to keep the salmon in equilibrium. Using the equation F(t) - W = Ma, we can solve for the normal force, which is the apparent weight of the salmon in this case. Plugging in the given values, we get F(t) = (5.0kg)(9.8m/s^2) - (5.0kg)(3.2m/s^2) = 36.0N. Therefore, the apparent weight of the salmon when the elevator is moving downward is 36.0N.