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Mechanics exam style question

username1550793

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(edited 3 years ago)

Reply 1

Notnek

Original post by tiger1296

Hello guys, I have a small question that I can't see to work out, would appreciate if anyone can help me work out the solution.

During, an engineer decides to take the lift going down, while standing on a scale placed on the lift’s floor. The engineer has a mass of 80 kg, and the lift is operating at an initial transient condition, with an acceleration of 2 m/s². Is the force indicated by the scale higher than, lower than or equal to the weight of the engineer??

The only forces acting on the man are gravity pulling him down and the reaction from the scale pushing him up. It's important to note that these are not a Newton's 3rd law pair since they act on the same object (the man). This means the forces aren't necessarily the same.

The man is accelerating down, which means that the two forces acting on him cannot be balanced i.e. the gravity force (the man's normal weight when not accelerating) must be greater than the force exerted on him by the scale.

Now by Newton's 3rd law, the force exerted by the man on the scale (this will be the reading on the scale) must be equal to the force exerted by the scale on the man.

Try to use all of the information to answer the question. A diagram will help.

Reply 2

Notnek

Original post by tiger1296

So I need to work out the force of the lift and the force of the man inside the lift?

You don't need to work out anything. You need to read my post above carefully and make sure you understand everything I said - please let me know if there's anything that doesn't make sense. If you understand everything in the post then you should be able to see the answer.

And you should have a diagram of the forces acting on the man, if you haven't drawn one already.

Reply 3

username1550793

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(edited 3 years ago)

Reply 4

Notnek

Original post by tiger1296

I do need to show calculations to justify whatever answer I conclude

Okay start with a force diagram of the man (show the acceleration of 2m/s^2 downwards). The force on him acting down will be 80g. The force acting on him up (force exerted by the scale) is unknown. Resolve the forces and use F = ma to find the force acting up. Please post your working / thoughts if you get stuck.

Reply 5

Notnek

Original post by tiger1296

What do you mean resolve the forces? I have drawn a diagram on paper, labelled it like you said. Do I multiply the mans weight by the acceleration?

You will have done this process before many times.

E.g. you have a car of weight 1.5kg being pulled with a 20N force and the resistance is 5N. What is the acceleration?

So you do Resultant force = ma

-> 20 - 5 = 1500a

You need to do the same thing here with your two forces and acceleration. The acceleration is downwards so it would be a good idea to take downwards as positive.

Reply 6

Notnek

Original post by tiger1296

I am so confused now, I don't even know what two forces you're referring to, this is really bad.

Did you understand the car example?

I'll start the question for you:

The two forces acting on the man in the lift are the force due to gravity acting downwards with magnitude 80g and the reaction of the scale on the man which I'll call R. These should be the two forces in your diagram.

Taking downwards as positive, the resultant force of the man is 80g - R.

So using F = ma you get

80g - R = 80 x 2

Does this make sense?