What Is the Minimum Hanging Mass Required to Keep a Block Stationary on a Wedge?

[derosef55]

1. A 1 kg wedge is pulled across a rough floor (uk=0.20) via contact with a 2 kg block as shown below. The shape of the wedge is that of an isosceles right triangle. a rope passes from the 2 kg block and over a mass-less and friction-less pulley and is attached to a hanging mass. Find the minimum value of the hanging mass so that the 2 kg block remains at rest on the inclined surface of the 1 kg block as it accelerates. The coefficients of friction between the 1 kg block and he 2 kg block are us=0.60 and uk=0.40. 2. Newton's Laws of Motion3. We did no problems like this in class before, so I'm pretty stumped on how to solve it, but I got a final answer of 2.3 kg for the hanging mass. Can anyone help? The picture for the problem is shown below.

 

[PeroK]

How did you get 2.3kg?

It seems to me that when the unknown mass reaches a certain value, the pair of blocks will start to move together. Is this the mass you need to find?

And, if the mass gets larger then at another value the 1 kg block will start to slip. It doesn't seem like you're being asked to find this.

Does that fit your interpretation of what you have to find?

 

[derosef55]

It seems to me that when the unknown mass reaches a certain value, the pair of blocks will start to move together. Is this the mass you need to find?

Yes, I need to find the mass of the hanging block such that it causes the other two blocks to move together, but NOT cause the top block (the one the rope is connected to, the 2 kg one) to move. Hopefully that makes more sense now lol sorry I should have clarified that more.

 

[PeroK]

Yes, I need to find the mass of the hanging block such that it causes the other two blocks to move together, but NOT cause the top block (the one the rope is connected to, the 2 kg one) to move. Hopefully that makes more sense now lol sorry I should have clarified that more.

What have you worked out so far? You said you got 2.3kg. How did you get that?

 

[derosef55]

Well, first I made the force diagrams and noted the action/reaction pairs. I'll put them on so you can see if I did those right or not. Then, I figured the net force in the y-direction for the 2 kg block had to be equal to 0, so I solved for the Force of block one on block two (F1on2). Then, I figured that the net force in the x-direction for the 2 kg block had to be 0 (I rotated the coordinate axes for that block) in order for it not to move, so using the F1on2 I just found, I found the static friction on that block by multiplying it by the coeff. of static friction given in the problem, and solved through for the tension in that block, then used that as the tension in the hanging mass as well because they both have tension forces coming from the same rope, and I set that equal to the "mg" force in the hanging mass and solved for m and got 2.3 kg. I guess that seems like a reasonable answer, but I feel like it couldn't have been that simple, there's so many other aspects to this problem.

 

[derosef55]

here's my full page of work so far.

 

[PeroK]

here's my full page of work so far.

I can't really work out from your notes what exactly youi've done, but 2.3 kg is too much. Here's why:

Unless the 2 kg block starts to slip, then effectively it's stuck to the 1 kg block and the internal forces (between the blocks) must be equal and opposite. So, you consider them as a single block.

The maxiumum friction force of the two blocks and the ground is 3g*0.2 = 0.6g. That's all it will take to move them. So, a rough guess at the answer is about 1 kg.

The key is to work out the minimum and maximum masses for which the 2kg block does not slip. Then, within that range, you can look at the external forces.

 

[derosef55]

Unless the 2 kg block starts to slip, then effectively it's stuck to the 1 kg block and the internal forces (between the blocks) must be equal and opposite. So, you consider them as a single block.

So what you're saying is I should treat the two blocks as one?
And what do you mean by the "internal forces between the blocks"? Like the static friction between them?

 

[PeroK]

So what you're saying is I should treat the two blocks as one?
And what do you mean by the "internal forces between the blocks"? Like the static friction between them?

Yes. Imagine the following, on a flat horizontal surface:

You have a small block touching the ground and a larger block on top of this - and quite a lot of friction between them. You start to push the larger block. Unless the two blocks start to slip, it's effectively just a single body problem. You only need to consider external forces.

You can use the same principle in your problem - although it's much more complicated.

 

[derosef55]

The maxiumum friction force of the two blocks and the ground is 3g*0.2 = 0.6g. That's all it will take to move them. So, a rough guess at the answer is about 1 kg.

The key is to work out the minimum and maximum masses for which the 2kg block does not slip. Then, within that range, you can look at the external forces.

So what about this part then? I'm not sure I follow what you're saying in the first part of this about the max friction force.

 

[PeroK]

So what about this part then? I'm not sure I follow what you're saying in the first part of this about the max friction force.

If the blocks were stuck together and you tried to pull them horizontally, then that's the force you would need to overcome the maximum frictional force with the ground.

I'm going offline, but here are a few ideas:

First, note that on their own, the 2kg block will slide down the incline. So, there's a minimum mass needed to stop that. I'd calculate that first to get a feel for the problem. The answer is ##0.4 \sqrt{2} \ kg## See if you can get that.

Second, find the mass needed to make the 2kg block slide up the slope. The answer is ##1.6 \sqrt{2} \ kg## See if you can get that.

It's not really necessary to calculate these two masses, although it confirms the problem makes sense and helps validate the answer.

What I'm saying is that between these two masses, the blocks don't slide, so you can look at the external forces only.

So, you can try solving the problem using only external forces. And hope the answer comes out between the two above!

PS I guess that ##1.6 \sqrt{2} = 2.3## is the answer you got! That's what's needed to pull the block up the slope.

 

[derosef55]

Ok I finally solved for the hanging mass needed to keep the 2 kg block from not moving, but I got .56 kg. Then I used that to find the net force in the x-direction for the whole system and got -1.2 N. So I'm thinking something went wrong somewhere lol. Basically, I don't know what to do after finding that minimum mass. I just plugged it into a bunch of stuff to see what I could come up with anywhere but it led me to a negative net force in the x-direction for the system somehow, so something somewhere is not right.

 

[derosef55]

PS I guess that 1.62√=2.31.6 \sqrt{2} = 2.3 is the answer you got! That's what's needed to pull the block up the slope.

oh I just read this last part so my answer was correct then? It does fall between those two values, but that's the highest value in that range you gave me. The problem asks for the "minimum value" of the hanging mass so the 2 kg block remains still as the system moves.

...so confused lol

 

[derosef55]

So, there's a minimum mass needed to stop that. I'd calculate that first to get a feel for the problem. The answer is 0.42√ kg0.4 \sqrt{2} \ kg See if you can get that.

Second, find the mass needed to make the 2kg block slide up the slope. The answer is 1.62√ kg1.6 \sqrt{2} \ kg See if you can get that.

how did you get this range anyways?

 

[Jazz]

PS I guess that ##1.6 \sqrt{2} = 2.3## is the answer you got! That's what's needed to pull the block up the slope.

Hi PeroK,

I've been following the thread, but here I got stuck in this part (I hope not to confuse to derosef55 with my doubt).

If there is a downward force that is equal to ##\small{w_{(2kg)}\sin(\theta)}##. A tension with this magnitude but upwards is needed to prevent the block from going down the slope. If I don't want the block to slip upwards, the 'extra-force' that the unknown mass can exert is ##\small{\mu_{s(12)}w\cos(\theta)}##. Then:

Note: ##\small{\mu_{s(12)}}## as the coefficient between the block and the triangle.

##F_{net} = F_{exerted} - [w_{(2kg)}\sin(\theta) + \mu_{s(12)}w\cos(\theta)]##

##0 = F_{exerted} - [w_{(2kg)}\sin(\theta) + \mu_{s(12)}w\cos(\theta)]##

##F_{exerted} = w_{(2kg)}\sin(\theta) + \mu_{s(12)}w\cos(\theta)##

If one does the math there, will find that the force exerted (that of the tension) is ##\small{\approx 22.17\ N}##, which should be supplied by a mass of ##\small{\frac{22.174\ N}{9.8\ m/s^2} = 2.26\ kg \approx 2.3\ kg}##.

Why can't a 2.3-kg mass be the higher part of the threshold that makes the triangle slip the floor without moving the other block?

 

[PeroK]

Hi PeroK,

I've been following the thread, but here I got stuck in this part (I hope not to confuse to derosef55 with my doubt).

If there is a downward force that is equal to ##\small{w_{(2kg)}\sin(\theta)}##. A tension with this magnitude but upwards is needed to prevent the block from going down the slope. If I don't want the block to slip upwards, the 'extra-force' that the unknown mass can exert is ##\small{\mu_{s(12)}w\cos(\theta)}##. Then:

Note: ##\small{\mu_{s(12)}}## as the coefficient between the block and the triangle.

##F_{net} = F_{exerted} - [w_{(2kg)}\sin(\theta) + \mu_{s(12)}w\cos(\theta)]##

##0 = F_{exerted} - [w_{(2kg)}\sin(\theta) + \mu_{s(12)}w\cos(\theta)]##

##F_{exerted} = w_{(2kg)}\sin(\theta) + \mu_{s(12)}w\cos(\theta)##

If one does the math there, will find that the force exerted (that of the tension) is ##\small{\approx 22.17\ N}##, which should be supplied by a mass of ##\small{\frac{22.174\ N}{9.8\ m/s^2} = 2.26\ kg \approx 2.3\ kg}##.

Why can't a 2.3-kg mass be the higher part of the threshold that makes the triangle slip the floor without moving the other block?

Yes, that's right. At 2.3 kg, the 2kg block will start to slide up the slope.

 

[PeroK]

Ok I finally solved for the hanging mass needed to keep the 2 kg block from not moving, but I got .56 kg. Then I used that to find the net force in the x-direction for the whole system and got -1.2 N. So I'm thinking something went wrong somewhere lol. Basically, I don't know what to do after finding that minimum mass. I just plugged it into a bunch of stuff to see what I could come up with anywhere but it led me to a negative net force in the x-direction for the system somehow, so something somewhere is not right.

The 0.56 kg is correct (that's ##0.4 \sqrt{2}##).

To solve the actual problem, you have two main external forces: gravity and tension from the block. You need to calculate the nett force on the block: you need to express this force in horizontal and vertical components.

Then, you can calculate the friction with the surface and work out at what point the blocks start to move.

 

[Jazz]

Yes, that's right. At 2.3 kg, the 2kg block will start to slide up the slope.

But the reason is not clear to me yet |: . If my system is the 2-kg block, then having the forces that I wrote above would be enough to keep it in balance (preventing it from slippering). Then If the 1-kg triangle is taken to be the system, it has the horizontal component of the tension and the horizontal component of wcos(theta) acting toward the right, making it to move toward that direction.

 

[PeroK]

But the reason is not clear to me yet |: . If my system is the 2-kg block, then having the forces that I wrote above would be enough to keep it in balance (preventing it from slippering). Then If the 1-kg triangle is taken to be the system, it has the horizontal component of the tension and the horizontal component of wcos(theta) acting toward the right, making it to move toward that direction.

Yes, that's right. Below 2.3 Kg, both blocks move together to the right. But, at 2.3 Kg the 2 Kg block starts to move and the system changes: the two blocks no longer move together uniformly.