TBRH, Physics, Passage 1

[Sammy1024]

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1. What is the tension in the cord in Trial 3 of Experiment 2?

Trial 3: M = 3kg, m = 1500g, a = g/6

The picture is of a table top:
------------| (like that)
|

and then there's a block with mass M on top of the table top, and then a pulley at the edge of the table top, with mass m hanging over and the string is 50 cm from the edge of the table, with a total length of 75 cm.

I thought this would be a simple T = mg sort of thing but apparently not.

I did T = mg since the little block is hanging off the table, so T = 1.5kgx10, for 15 N but the answer is 12.5 N.

2. Before trial 1 of experiment 1 is complete, when the total mass m hanging off the table is just 200 g, what is the force of static friction on the wooden block?

Trial 1: M = 1 kg, Threshold mass = 396g. coefficient of static friction is 0.4

I tried using the Fs = us (static friction) x F, and then replaced F with m x g but I got 0.8 and the answer is 2N. (I used mg instead of a because gravity is a form of acceleration and thought it could replace the a)

Where did I go wrong?

 

[inasensegone]

1. What is the tension in the cord in Trial 3 of Experiment 2?

Trial 3: M = 3kg, m = 1500g, a = g/6

The picture is of a table top:
------------| (like that)
|

and then there's a block with mass M on top of the table top, and then a pulley at the edge of the table top, with mass m hanging over and the string is 50 cm from the edge of the table, with a total length of 75 cm.

I thought this would be a simple T = mg sort of thing but apparently not.

I did T = mg since the little block is hanging off the table, so T = 1.5kgx10, for 15 N but the answer is 12.5 N.

2. Before trial 1 of experiment 1 is complete, when the total mass m hanging off the table is just 200 g, what is the force of static friction on the wooden block?

Trial 1: M = 1 kg, Threshold mass = 396g. coefficient of static friction is 0.4

I tried using the Fs = us (static friction) x F, and then replaced F with m x g but I got 0.8 and the answer is 2N. (I used mg instead of a because gravity is a form of acceleration and thought it could replace the a)

Where did I go wrong?

There is no TBRH... there is a TPRH, however I checked my passage 1 in physics and this is not the same passage. I have the 2010 science workbook, which I was told is the same as the current one. Can you clarify? If I have the passage, I'll check it out.

 

[Sammy1024]

Oh sorry, mistyped the title. I have the 2010 one. But it's the in class-compendium book.

 

[Sammy1024]

I posted a picture of the passage in the first post.

 

[inasensegone]

View attachment 179757

1. What is the tension in the cord in Trial 3 of Experiment 2?

Trial 3: M = 3kg, m = 1500g, a = g/6

The picture is of a table top:
------------| (like that)
|

and then there's a block with mass M on top of the table top, and then a pulley at the edge of the table top, with mass m hanging over and the string is 50 cm from the edge of the table, with a total length of 75 cm.

I thought this would be a simple T = mg sort of thing but apparently not.

I did T = mg since the little block is hanging off the table, so T = 1.5kgx10, for 15 N but the answer is 12.5 N.

2. Before trial 1 of experiment 1 is complete, when the total mass m hanging off the table is just 200 g, what is the force of static friction on the wooden block?

Trial 1: M = 1 kg, Threshold mass = 396g. coefficient of static friction is 0.4

I tried using the Fs = us (static friction) x F, and then replaced F with m x g but I got 0.8 and the answer is 2N. (I used mg instead of a because gravity is a form of acceleration and thought it could replace the a)

Where did I go wrong?

Ahhh... I think I see where you went wrong...

1. The question is almost as simple as T = mg... but you need to factor in the acceleration. The big block M is accelerating to the right at *a = g/6*. This means that the hanging mass m is also accelerating downward at a rate of g/6. When you calculate the tension, you must take this acceleration into account. The only way that the tension would simply be equal to mg, is if both blocks were static. In this case they are moving so the full tension of 15 N does not apply. The tension is instead given as follows: Tension = mg - ma. Remember that a = g/6. This becomes T = 1.5(10) - 1.5(10/6). This reduces to T = 15 - (15/6). Fifteen divided by 6 is equal to 2.5, so the tension is 12.5 N.

2. OK for this one it is important to remember that static friction only provides enough force to prevent slippage... up to a certain point. That means it provides enough force to oppose sliding, sort of as a reactionary force. For example, if you had a large box on the floor and you pushed it lightly and it didn't move... this is because static friction provided a force opposite to your pushing force. If you pushed harder and it still didn't move, this is because the static force similarly increased. It simply matches the amount of force... Newton's 3rd law. An equal and opposite force... up to a point. Beyond that point, you have exceeded the maximum amount of force that static friction can provide. This is why you always see static friction defined as Force of static friction is less than or equal to µN. If the force of static friction were to somehow exceed the force that you apply, then the box would somehow push you backward as you pushed on it. OK, so anyway, to answer your question, the force of static friction on the wooden block is simply equivalent to the weight of the hanging block. F=mg = 0.2*10 = 2N. The reason is because in trial 1 the threshold mass is given as 396 grams. Well we only have 200 grams, so we haven't exceeded the threshold mass that is needed to produce sliding. This means that the static friction will supply an equal and opposite force sufficient to prevent sliding.

I hope that makes sense. Let me know if it doesn't.

 

[DrknoSDN]

if TLDR you can always easily determine tension in an accelerating object by finding the fraction of acceleration.

Like inasensegone said, tension = mg when everything is static.
Here The difference between gravity and acceleration is 5/6 so you just times the static tension by that... 1.5kg * 10N/kg * (5/6) = 12.5

Example2: if a = 2/3 g, you just do 15N * (1/3) = 5N tension

I suppose you assume that the mass on the table is stationary and just reduce gravity by the acceleration of the table mass.