# Do all machines decrease force by a factor of 2?

#### Halcyon32

2+ Year Member
I did a practice question that said a mass is on a board which pivots at one end and at the other end, a pulley is attached to the board. So 2 machines are acting on the mass. Then the solution said that both the lever and pulley system reduce the force by a factor of 2 each.

Even if you don't understand the setup of the question, my question is: is it just something we need to know that every machine decreases work by a factor of 2, or was there something in this problem that i missed that indicated this?

Nothing like that was outright stated by the problem as far as I could tell.

#### popopopop

7+ Year Member
What's the question? Short answer is no.

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2+ Year Member
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2+ Year Member

#### popopopop

7+ Year Member
Ok, so work = force x distance. You can either increase force or increase distance to increase your work output. That's the mechanical advantage of a pulley.

So we know the force from the mass is roughly 100 N. If F1= mg and we have 2 tensions pulling up balancing the block, then F2 (pulley force) = 2T. If F2 = F1, which we want it to be, we can say 2T = mg.

If 2T = mg or mg - 2T = 0, then T = mg/2 = F2. Our pulley force is 50 N in this case because we have 2 tensions. This is true in the case that if F2 is pulling straight up, opposing the mass.

However, our force is pulled at an angle(sin is vertical component) so we account that too with F2= mg(sin30)/2 = 25 N. The vertical component must equal to the hanging weight and all of these components add together to give a net force of 0, static equilibrium.

I tried to make it clear lol.

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#### Halcyon32

2+ Year Member
Ok, so work = force x distance. You can either increase force or increase distance to increase your work output. That's the mechanical advantage of a pulley.

So we know the force from the mass is roughly 100 N. If F1= mg and we have 2 tensions pulling up balancing the block, then F2 (pulley force) = 2T. If F2 = F1, which we want it to be, we can say 2T = mg.

If 2T = mg or mg - 2T = 0, then T = mg/2 = F2. Our pulley force is 50 N in this case because we have 2 tensions. This is true in the case that if F2 is pulling straight up, opposing the mass.

However, our force is pulled at an angle(sin is vertical component) so we account that too with F2= mg(sin30)/2 = 25 N. The vertical component must equal to the hanging weight and all of these components add together to give a net force of 0, static equilibrium.

I tried to make it clear lol.
But the question isn't asking for it to be in static equilibrium, it's saying to take it out of equilibrium by lifting the mass. Or maybe I'm understanding it wrong? And why didn't you factor in the lever system? Also, the solutions manual says that the angle the force is acting is irrelevant, just that the lever and pulley system reduce the force by a factor of 2 each, which is the part I really don't understand. Any ideas?
I really appreciate your help by the way, thanks a lot!

#### popopopop

7+ Year Member
So taking out equilibrium to raise it would mean you need a force of 25.1 or something like that, which is close to 25. You can't assume that the starting point is 25 because well, you don't know what the force is to keep it in static equilibrium. To answer these problems, you get a force diagram going and set out to make a net force equation first. I did factor in the lever system, that's why I was mentioning the two strings having essentially two tensions pulling up on the plank, which is true regardless of the angle . Tension is constant in the string yes, but I don't know why angle doesn't matter. I'll be googling the heck out of this to find the answer lol.

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#### Halcyon32

2+ Year Member
So taking out equilibrium to raise it would mean you need a force of 25.1 or something like that, which is close to 25. You can't assume that the starting point is 25 because well, you don't know what the force is to keep it in static equilibrium. To answer these problems, you get a force diagram going and set out to make a net force equation first. I did factor in the lever system, that's why I was mentioning the two strings having essentially two tensions pulling up on the plank, which is true regardless of the angle . Tension is constant in the string yes, but I don't know why angle doesn't matter. I'll be googling the heck out of this to find the answer lol.
Ok, that clears a lot of things up. So I shouldn't assume that all machines decrease force by a factor of 2? And I think the reason that the angle doesn't matter is because work is independent of the angle. You can solve 2D kinetics problems by using conservation of energy rather than splitting it up into x and y components and using the kinematics equations. And since machines act to decrease force required for work, I guess that means you don't need the angle. But I haven't even really convinced myself with that explanation and I'll definitely be looking into it, as well. Thanks for your help and please let me know if you come to any sort of revelation about the problem

#### popopopop

7+ Year Member
Yah, I'm trying to find another similar question. Pulley system depends on how many pieces of strings supports the weight. If you have 4 pieces of strings acting up, then mg/4 = F because mg - 4F = 0.

It kills me a bit because I did so many of these in physics 1, but I'm not 100% on the concept yet since it's been so long. I thought it was 50 at first if angle doesn't matter. Any chance the book is wrong? lol

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#### Halcyon32

2+ Year Member
Yah, I'm trying to find another similar question. Pulley system depends on how many pieces of strings supports the weight. If you have 4 pieces of strings acting up, then mg/4 = F because mg - 4F = 0.

It kills me a bit because I did so many of these in physics 1, but I'm not 100% on the concept yet since it's been so long. I thought it was 50 at first if angle doesn't matter. Any chance the book is wrong? lol
Ok, I figured it out! We established that the pulley decreases the force by a factor of 2 based on your previous explanation. Now, the reason the lever system also decreases the force by a factor of 2 is because the lever arm of the string where it attaches to the right end of the board is twice that of the lever arm of the mass sitting in the center of the board, thus:
F(in)*d(in)=F(out)d(out)---->2F(in)=F(out)=mg=100----->F(in)=50N. So, since we applied the force of the pulley on the lever twice as far as the mass is acting on the lever, the applied force on the lever is half of the force of the mass (which is just its weight). This on top of the added advantage of the pulley itself decreases the force by a factor of 4 total, down to 25N
In regards to the angle: Whatever angle the rope is being pulled at, the tension is still gonna be the same throughout (pull at a force of 25N at an angle of 75 degrees and the tension in the rope is still going o be 25N throughout, assuming of course the rope is massless and the pulley is frictionless, which is a necessary assumption) and since the part that of the pulley that we see is directly acting on the system are the two vertical pieces of rope, we don't need to worry about the angle! Let me know if this made sense and seems correct because damn it took me a long time to figure out.

#### popopopop

7+ Year Member
^I thought of that for a brief moment thinking that it did say the mass was in the center. A friend of mine suggested something else and I never went with it. So basically the pulley system is pulling 50N up if we didn't factor in the lever arm. Because we are pulling on a plank with a lever arm, torque/force (in) must = torque/force(out) as you said. The distance where the pulley attaches doubles it's r/torque in this case, so our force is reduced by 2 to 25N.

We didn't get it right away, but working it out was useful. Now I know to ignore the angle.

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#### Halcyon32

2+ Year Member
Yep, gained some good insight on the subject. Thanks again for your help