# Supposedly very easy questions dealing with projectiles, equilibrium, etc.

1. Nov 10, 2007

### gamerzgamerz

even if you only have time to do one, i'd really appreciate it!

A ball is thrown with an initial speed of 15 m/s at an angle of 53 degrees above the horizontal from the top of a 35 m building. if g = 9.8 m/s^2 and air resistance is negligible, then what is the magnitude of the horizontal components of velocity as the rock strikes the ground?

(the answer is 9.0 m/s, but i have no idea how to get to that.)

A 500-N tightrope walker stands at the center of the rope such that each half of the rope makes an angle of 10 degrees with the horizontal. what is the tension in the rope?

(answer: 1440 N, but i somehow got 2900 N)

a vault is opened by applying a force of 300 N perpendicular to the plane of the door, 0.80 m from the hinges. find the torque due to this force about an axis through the hinges.

2. Nov 10, 2007

### hage567

You need to show some effort in working on these questions. Can you show what equation might be used? What have you tried? Can you explain more specifically where you are stuck?

For the first one, what kinematic equations might be useful given the information in the question? Do you know how to resolve the initial velocity into its components?

For the second, show what you did that got you the wrong answer.

For the third, what's the definition of torque?

3. Nov 10, 2007

### Vidatu

Nuts, hage567 beat me to the reply. Oh well; I can still contribute.

Draw a diagram for each problem. Then, when solving, make sure you address every part of that diagram. At least one of your problems came from forgetting some part of the system.

4. Nov 10, 2007

### catkin

Let's start with the one you tried. How did you get 2900 N? If we help you find out how to do that correctly you may get insight enough to have a go at the other two.

5. Nov 10, 2007

### gamerzgamerz

the first one: i only know how to do a projectile from the ground or simply dropping it from on top of a cliff. i'm not sure about ones with an indicated angle.
the 2nd one: i drew a diagram and separated the tension into T1 on the right side and T2 on the left and found out T1 = T2 so 2(T1)sin10=500 --> T1=1440 and so i multiplied it by 2 and got 2900 N.
the third: i really don't know...

6. Nov 11, 2007

### Vidatu

For A), use trig to split the velocity into horizontal and vertical components.

For B), you were mostly right. How many times did you double the force?

For C), A moment/torque is defined as the product of a force and the perpendicular distance between that force and the point of interest.

7. Nov 11, 2007

### aq1q

ya... part a) is explained.. clearly by many of the ppl here.. for part b) draw a free body diagram.. that will do wonders.. the person isn't accelerating down.. so the weight must equal to the tension in both string.. therefore u don't need to multiply by 2.

and for part c) t= r x F sin(theta) (since it says that its perpendicular.. sin90=1) (t- torque, and F - force)

8. Nov 11, 2007

### gamerzgamerz

Thanks for the help! Ok i got question 3.
i still don't understand why i don't have to multiply it by two for question 2.

9. Nov 11, 2007

### aq1q

Because there aren't 2 ropes.. its still 1 rope..so the force is shared evenly between the two segments of the ropes. So u just solve for T. If you still don't get it, i can draw it out.. but I am not even sure how to put pictures here

Last edited: Nov 11, 2007
10. Nov 11, 2007

### gamerzgamerz

oh please do if you don't mind. you can upload it on imageshack.us and afterwards it'll give you the code you need to copy and paste it on a forum.
thanks in advance if you get the chance to do it.

11. Nov 11, 2007

### aq1q

12. Nov 11, 2007

### Vidatu

Each side of the rope supports half the weight, and as tension in a rope is constant throughout, the tension in each rope is the overall tension.

This is for aq1q, on how to upload images:

First, go to photo bucket, and get the item link.

Next, you'll click on the Add Image button