What Is The Mass Of Block B In This Specific Example?

[Jacques]

 

(A question in grade 11 Physics paper.)

 

Information given:

 

Two blocks are connected to each other with a rope that runs over a frictionless pulley.

 

The system is stationary/at rest.
Coefficient of static friction between the 5.1kg block and slope surface is 0.3.
The slope has an inclination of 27 degrees.

 

Question: What is the mass of block B?

 

My answer:

 

First calculate maximum static friction of 5.1kg block:
fs(max) = Fn * μs
fs(max) = (5.1 * 9.8 * cos27) * 0.3
fs(max) = 13.3598...N

 

Then calculate the gravitational force of the 5.1kg block parallel to the slope
Fg(parallel) = Fg * sin27
Fg(parallel) = (5.1 * 9. 8) * sin27
Fg(parallel) = 22.6904...N

 

Now, since they didn't tell us whether the block is on the brink of moving or which direction the static friction is working in, I will calculate the min and max weight of block B:

 

If the static friction works down the slope (suppose the force from the rope on the 5.1kg block is greater than Fg(parallel) of the 5.1kg block), one can get the max weight of B:
Max tension in rope = 22.6904... + 13.3598... ≈ 36.0502N
Thus max weight ≈ 36.0502 / 9.8 ≈ 3.6786kg

 

If the static friction works up the slope (suppose the force from the rope on the 5.1kg block is less than Fg(parallel) of the 5.1kg block), one can get the min weight of B:
Min tension in rope = 22.6904... - 13.3598... ≈ 9.3306N
Thus min weight ≈ 9.3306 / 9.8 ≈ 0.9521kg

 

Am I correct if I say the mass of block B is anything from 0.9521kg to 3.6786kg?
I am asking because our memorandum says the mass of B is simply 0.9521kg and nothing else. And I don't want to lose around 8 marks...

Edited November 26, 2017 by Jacques

[Maine farmer]

 

 

 

 

 

 

Now, since they didn't tell us whether the block is on the brink of moving or which direction the static friction is working in, I will calculate the min and max weight of block B:

 

 

Well, they told you the system is stationary, which means that all forces must balance each other out.

[exchemist]

Well, they told you the system is stationary, which means that all forces must balance each other out.

Yes but that is possible for a range of mass B values, due to the friction of the 5.1kg mass with the slope. The friction force can be anywhere between a force of 0.3 x 5.1 x g acting on the block in the up-slope direction or the same acting in the down-slope direction.

 

So long as the sum of the tension in the string and this range of friction force equals the component of the block's weight in the down-slope direction, the block will not move.

 

Er, I think. :)

[Maine farmer]

Yes but that is possible for a range of mass B values, due to the friction of the 5.1kg mass with the slope. The friction force can be anywhere between a force of 0.3 x 5.1 x g acting on the block in the up-slope direction or the same acting in the down-slope direction.

 

So long as the sum of the tension in the string and this range of friction force equals the component of the block's weight in the down-slope direction, the block will not move.

 

Er, I think. :)

I just don't think the creators of the problem intended the problem to be that complicated.  I would have simply calculated the minimum force for the tension in the string.  Any more force, and the block would be pulled up the slope - any less and it would slide down the slope.  Or is that my bias for simple solutions?

[exchemist]

I just don't think the creators of the problem intended the problem to be that complicated.  I would have simply calculated the minimum force for the tension in the string.  Any more force, and the block would be pulled up the slope - any less and it would slide down the slope.  Or is that my bias for simple solutions?

OK, the minimum would imply the question requires the minimum mass to just prevent the block sling down. Whereas the way it was expressed in the OP it could be taken to be either that or the maximum mass that just avoids it sliding up. 

 

By the way there was an error in my formula yesterday: the max friction should have been 0.3 x 5.1kg x g x Cos 27deg (the component of weight normal to the plane). I'm always like this with mechanics - keep leaving things out and having to go back and correct them.

 

 

Edited November 28, 2017 by exchemist

[exchemist]

 

 

 

(A question in grade 11 Physics paper.)

 

Information given:

 

Two blocks are connected to each other with a rope that runs over a frictionless pulley.

 

The system is stationary/at rest.

Coefficient of static friction between the 5.1kg block and slope surface is 0.3.

The slope has an inclination of 27 degrees.

 

Question: What is the mass of block B?

 

My answer:

 

First calculate maximum static friction of 5.1kg block:

fs(max) = Fn * μs

fs(max) = (5.1 * 9.8 * cos27) * 0.3

fs(max) = 13.3598...N

 

Then calculate the gravitational force of the 5.1kg block parallel to the slope

Fg(parallel) = Fg * sin27

Fg(parallel) = (5.1 * 9. 8) * sin27

Fg(parallel) = 22.6904...N

 

Now, since they didn't tell us whether the block is on the brink of moving or which direction the static friction is working in, I will calculate the min and max weight of block B:

 

If the static friction works down the slope (suppose the force from the rope on the 5.1kg block is greater than Fg(parallel) of the 5.1kg block), one can get the max weight of B:

Max tension in rope = 22.6904... + 13.3598... ≈ 36.0502N

Thus max weight ≈ 36.0502 / 9.8 ≈ 3.6786kg

 

If the static friction works up the slope (suppose the force from the rope on the 5.1kg block is less than Fg(parallel) of the 5.1kg block), one can get the min weight of B:

Min tension in rope = 22.6904... - 13.3598... ≈ 9.3306N

Thus min weight ≈ 9.3306 / 9.8 ≈ 0.9521kg

 

Am I correct if I say the mass of block B is anything from 0.9521kg to 3.6786kg?

I am asking because our memorandum says the mass of B is simply 0.9521kg and nothing else. And I don't want to lose around 8 marks...

 

I think you have given an excellent answer, if the problem is presented exactly as you have described, i.e. without specifying whether a minimum or maximum mass is required. I do not think you will lose marks for pointing out that the mass of block B can be anything between those values.

[Maine farmer]

I think you have given an excellent answer, if the problem is presented exactly as you have described, i.e. without specifying whether a minimum or maximum mass is required. I do not think you will lose marks for pointing out that the mass of block B can be anything between those values.

For some reason this problem popped back into my head while doing chores, and I realized that the student was thinking more clearly than I.  I suppose I really do get fixated on finding the simplest solution.  The problem is an example of how imprecise wording can make things more complicated.

 

I hope the student got some extra points for the extra thought!