# In what case do we use the cos(theta) in W?

#### laczlacylaci

2+ Year Member

For here, I went for W=(40*10)*(5)*(cos60)=100J
The key just said to do 40*10*5=200J. So in what case do we use cos theta? I am assuming that they used cos(0)=1, but where is this zero degree from?

I understand that the angle should be between the F and d, but I don't think I can picture where to draw that angle. I would imagine that M is lifted up perpendicular to the dashed line, making the angle between F and D 30, but that would not lead me to the correct answer.

#### Kpw101

5+ Year Member
View attachment 207910
For here, I went for W=(40*10)*(5)*(cos60)=100J
The key just said to do 40*10*5=200J. So in what case do we use cos theta? I am assuming that they used cos(0)=1, but where is this zero degree from?

I understand that the angle should be between the F and d, but I don't think I can picture where to draw that angle. I would imagine that M is lifted up perpendicular to the dashed line, making the angle between F and D 30, but that would not lead me to the correct answer.
I may be wrong but it looks like you could have easily just calculated work done from the potential energy gained. If the mass is lifted 5 meters and weighs 4kg then (4)(5)(10) = 200J. For work done you use Cos(theta) only when the displacement and the force are not in the same direction. When displacement and force are in the same direction Cos(0) = 1 so it cancels out. Theta is the angle between force and displacement.

You are not given the force so you cannot use the equation F*Dcos(theta) = W. They simply calculate potential energy gained = Work done. PE = mgh.

OP

#### laczlacylaci

2+ Year Member
I may be wrong but it looks like you could have easily just calculated work done from the potential energy gained. If the mass is lifted 5 meters and weighs 4kg then (4)(5)(10) = 200J. For work done you use Cos(theta) only when the displacement and the force are not in the same direction. When displacement and force are in the same direction Cos(0) = 1 so it cancels out. Theta is the angle between force and displacement.

You are not given the force so you cannot use the equation F*Dcos(theta) = W. They simply calculate potential energy gained = Work done. PE = mgh.
As for displacement, isn't the displacement perpendicular to the dashlined? Making the angle between F and displacement 30?

#### DaWeasel

2+ Year Member
The work is being done on the 4 kg mass. The force is evaluated with respect to the object the work is being done on. The mass moves in the same direction as the applied force, or in parallel, so the angle between the applied force and the displacement in this case is zero.

OP

#### laczlacylaci

2+ Year Member
The work is being done on the 4 kg mass. The force is evaluated with respect to the object the work is being done on. The mass moves in the same direction as the applied force, or in parallel, so the angle between the applied force and the displacement in this case is zero.
I feel like I am still not understanding... Thanks for any clarification. This is what I have in mind. Why would the degree between them be 0?

#### Kpw101

5+ Year Member
I feel like I am still not understanding... Thanks for any clarification. This is what I have in mind. Why would the degree between them be 0?

View attachment 207914
You're missing the point I made. The equation used here is W = DeltaPE or change in potential energy. Potential energy = mgh so you don't need the angle at all. It is not in that equation. The equation for work is Fdcos(theta) = W. You are not given force and so are not suppose to use the equation. The angles are there to throw you off.

The work is being done on the 4 kg mass. The force is evaluated with respect to the object the work is being done on. The mass moves in the same direction as the applied force, or in parallel, so the angle between the applied force and the displacement in this case is zero.
The mass does not move in the same direction as the applied force. The force is applied 30 degrees relative to the pulley and so you would not be able to disregard cos(30) if you were actually given a force. But you are not given the value of the force in this problem so you do not use F*dcos(theta) = W and the direction the force is applied becomes irrelevant when solving the problem.

#### DaWeasel

2+ Year Member
I feel like I am still not understanding... Thanks for any clarification. This is what I have in mind. Why would the degree between them be 0?

View attachment 207914
The 60 angle in the figure is not relevant to solving the problem. Think about the set up here, with respect to the mass, not the F and the 60 in the figure. When the rope is pulled, the mass experiences a force directly upwards and moves in the same direction as this force. Since the mass moves in the same direction as the force it experiences, the angle in this case is zero.

#### Kpw101

5+ Year Member
The 60 angle in the figure is not relevant to solving the problem. Think about the set up here, with respect to the mass, not the F and the 60 in the figure. When the rope is pulled, the mass experiences a force directly upwards and moves in the same direction as this force. Since the mass moves in the same direction as the force it experiences, the angle in this case is zero.
Still incorrect here. I don't know how you would use the F*dCos(theta) = W equation if you were not given a number for the force to begin with. The equation used here is gain in potential energy by the object equals work or PE(gained) = Work = Mass*gravity*height. The force is not going in the same direction as the displacement. Even if it is a pulley system you cannot disregard the angle of the force.

Think about this if you were using a pulley, would it be easier to pull something up if you were pulling straight down or at an angle?

#### DaWeasel

2+ Year Member
Still incorrect here. I don't know how you would use the F*dCos(theta) = W equation if you were not given a number for the force to begin with. The equation used here is gain in potential energy by the object equals work or PE(gained) = Work = Mass*gravity*height. The force is not going in the same direction as the displacement. Even if it is a pulley system you cannot disregard the angle of the force.

Think about this if you were using a pulley, would it be easier to pull something up if you were pulling straight down or at an angle?
Yes, this problem is most easily solved using conservation of energy. I was answering the OPs question about where the 0 comes from. In F*dCos(theta) = W as applied to the mass, the theta is zero because the mass experiences displacement in the direction of the force applied to it. You do have a force here, the force needed to counteract gravity, and the displacement is given as 5. So the force is g, 10, the displacement is 5, and the mass is 4. 4 * 10 * 5 = 200.

laczlacylaci

#### Kpw101

5+ Year Member
I'm confused here. The force can't be equal to g. The value of 10 for gravity is acceleration not a force. Only by multiplying gravity by a mass do you get force because F = ma. What you're saying is you're multiplying force * displacement * mass. You're actually doing mgh which is still conservation of energy.
Yes, this problem is most easily solved using conservation of energy. I was answering the OPs question about where the 0 comes from. In F*dCos(theta) = W as applied to the mass, the theta is zero because the mass experiences displacement in the direction of the force applied to it. You do have a force here, the force needed to counteract gravity, and the displacement is given as 5. So the force is g, 10, the displacement is 5, and the mass is 4. 4 * 10 * 5 = 200.

#### DaWeasel

2+ Year Member
I'm confused here. The force can't be equal to g. The value of 10 for gravity is acceleration not a force. Only by multiplying gravity by a mass do you get force because F = ma. What you're saying is you're multiplying acceleration * displacement * mass. Neither acceleration nor mass is included in the Force * distance equation.
You're right, I wrote that out wrong, but I was saying the F component is m * g. My bad.

#### Kpw101

5+ Year Member
You're right, I wrote that out wrong, but I was saying the F component is m * g. My bad.
I think we're both right here, we're just looking at it from two different angles. The mgh equation is just a derived form of F*d(cos0/180). Probably confused OP more than we helped him.

DaWeasel