What is the force in the y-direction given the force and angle?

[joess]

thanks in advance for any help

 

[geoffjb]

thanks in advance for any help

What do you need help with? You must show an attempt to solve each problem, along with highlighting where you are having difficulty. No one is going to do the problems for you.

 

[joess]

oh sorry...

okay for the first one i don't even know where to begin... you don't need to solve the problem but i just want to know how to get started

for the second i know that the total of the forces must be 0 because there's no acceleration. i know (or think) that the y-component of F is 53900 N but i can't figure out how to get the x-component because there's no acceleration. so for this i just want to know how to get the x-component.

 

[geoffjb]

oh sorry...

okay for the first one i don't even know where to begin... you don't need to solve the problem but i just want to know how to get started

for the second i know that the total of the forces must be 0 because there's no acceleration. i know (or think) that the y-component of F is 53900 N but i can't figure out how to get the x-component because there's no acceleration. so for this i just want to know how to get the x-component.

To start, draw a free body diagram and label all forces. Then, determine the x- and y-components of each force. Note that the question does not want the block to move, so the mass is in static equilibrium (v_{net}=0 \text{, } \Sigma F_{y} = 0 \text{, and } \Sigma F_{x} = 0).

For the second, you're correct that there is no net acceleration (a state of dynamic equilibrium). There are, however, forces involved, so draw a free body diagram. Once you get the trig figured out, and everything is in its x- and y-components, you can find the force of the rotor.

 

[joess]

Okay for the first I got that the force of gravity is 49N, the normal force is 49N as well, the force of friction is 23.03, and obviously there's the force from the cord. What I can't figure out is how to relate the mass of the ball to the force. It's not just as simple as msin32, right?

And for the second one I drew a free body diagram but the only forces I can think of is F and the force of gravity. And I can't figure out how to find the x-component of F... shouldn't it be 0 since there's no acceleration? But there can't not be acceleration since the helicopter is moving... are there any other forces in the x-direction?

 

[geoffjb]

Okay for the first I got that the force of gravity is 49N, the normal force is 49N as well, the force of friction is 23.03, and obviously there's the force from the cord.

That all seems good. The force from the cord is known as tension.

What I can't figure out is how to relate the mass of the ball to the force. It's not just as simple as msin32, right?

Looking at the free body diagram of the "unknown" mass in the middle, you should be able to identify three forces. Since the ball is not moving, it means all the forces must equal _____. Knowing this, how are the tensions of the cords involved?

And for the second one I drew a free body diagram but the only forces I can think of is F and the force of gravity.

Good.

And I can't figure out how to find the x-component of F... shouldn't it be 0 since there's no acceleration? But there can't not be acceleration since the helicopter is moving... are there any other forces in the x-direction?

The fact that the helicopter is in level flight tells you something about the sum of the forces in the y-axis.

The fact that you're given the angle between the force of the rotor and the vertical means the remainder of the problem is simply plugging values into equations.

 

[joess]

Alright, so the tension on both sides of the cord are equal right? But how do I find the tension?

 

[geoffjb]

Alright, so the tension on both sides of the cord are equal right? But how do I find the tension?

The tensions are not equal. What about them is?

 

[joess]

Oh... so the tensions in the x-axis are equal?? Or what?

And in the second question, I've got the y-axis forces but how do I get the x-axis forces?

 

[geoffjb]

Oh... so the tensions in the x-axis are equal?? Or what?

Yes!

And in the second question, I've got the y-axis forces but how do I get the x-axis forces?

If you got the y-axis forces, you must have used a trig function to find the y-component of the rotor force. Use another trig function to find the x-component.

 

[joess]

So in that case shouldn't the tension pulling on the block be mgsin32? But if I use that I still don't get the right answer..

And in the second one shouldn't the forces on the y-axis just be the weight and the y-component of the force, which is equal to the wieght?

 

[geoffjb]

So in that case shouldn't the tension pulling on the block be mgsin32? But if I use that I still don't get the right answer..

And in the second one shouldn't the forces on the y-axis just be the weight and the y-component of the force, which is equal to the wieght?

The mass is in static equilibrium, so all component forces to the left must equal those to the right, and all component forces down must equal those going up.

Since the only x-components are the friction force, the tension of the horizontal string, and the x-component of the string to the wall, you need to balance them. Remember that the block is not moving. What does that tell you about the relationship of its x-component forces?

Once you figure out the tension of the string to the wall, figure out its y-component. That y-component must balance the weight of the mass in order to keep it in static equilibrium.

When you say "in the second one", do you mean the second string or the second question?

 

[joess]

Okay I think I've got the first question, working on it now to see if I get the right answer.

And by second one I meant the second question.

 

[geoffjb]

Okay I think I've got the first question, working on it now to see if I get the right answer.

Good.

And by second one I meant the second question.

Then yes, your statement was correct. And since acceleration is also zero in the x-direction, the statement also applies that way (once you apply the correct forces).

 

[joess]

Alright, got the first question, thanks.

So for the second question I still can't figure out how to get the x-direction forces. Is the x-component of the rotor's force the only x-axis force?

 

[geoffjb]

Alright, got the first question, thanks.

So for the second question I still can't figure out how to get the x-direction forces. Is the x-component of the rotor's force the only x-axis force?

I just re-read the second question; you don't need the x-component. All you need is the force itself.

If F_{y} = F \cos(\theta), solve for F.

(Sorry if my direction misled you.)