Archie's Archery Depot

More Info On Release String:

When a person is first starting of in archery there are many things that they will have to learn. They will have to learn how to hold the bow, place the arrow on the bow, draw back, aim, and follow through. It might seem overwhelming to people who don't know anything about the sport, but it is not that bad. The most important thing that you can do is to make sure that you practice.

When you are learning how to aim then there are a few things you will want to remember to do to give you the best chance of hitting the target. You will want to make sure that you are releasing in the right way every time and that until you release, you should keep the drawn string at the maximum position.

There are lots of little tips and things to watch out for when you are drawing your arrow back. In fact, this is a very important stage when you are shooting. Any forward movement by the arrow before you release it, is called "Creeping." This is usually caused by a person trying to relieve some strain by letting the hand move forward on the draw. Most of the time this is because the bow is too powerful for that person. Every bow has a rating for the amount of pounds it takes to draw the string back. If it is too much for you then it is recommended to try another bow.

Make sure that you draw the bow string to a full stretch every time that you fire a shot. This will help you to know the flight characteristics of the arrow's flight time. It is important to know this because if you don't, then you are going to have arrows that are going different distances from the loss of speed out of the bow. If the arrow doesn't have as much speed as it could coming out of your bow your arrows are always going to hit low.

Even if you pull the bow to a Full Draw you are going to get some fatigue in your arm. This will mean that creeping will happen more often for you. When people are first starting they might not understand just how much energy and tension that they are going to need to hold the bow at a full stretch. Using a bow that you can use over and over is what you are going to want to use. Even if you can pull a bow's string back once or twice doesn't mean that you should use it. You will want to look for a bow that you will be able to repeatedly draw back over a course of a practice.

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Mechanics involving inextensible string and a peg?

Can someone please help me with this question:
Particles A and B, or masses 0.2kg and 0.3kg respectively, are connected by a light inextensible string which passes over a fixed smooth peg. The system is released from rest, with B at a height of 0.25m above the floor. B descends, hitting the floor 0.5s later. All resistances to motion may be ignored. When B hits the floor it comes to rest immediately, and the string becomes slack. Find the length of time for which B remains at rest on the ground before being jerked into motion again.

First you need to find how fast the particles are accelerating.

Force balance on A:

mA*a = T - mA*g where

mA is mass of A,
T is tension in string,
g is gravity acceleration

Force balance on B:

-mB*a = T - mB*g

(note that I use -mB*a since it is moving down while A is moving up)

Subtract equations and solve for a:

(mA+mB)*a = (mB-mA)*g

a = (mb-mA)*g/(mA+mB)

= (0.3kg-0.2kg)*(9.8m/s^2)/(0.3kg+0.2kg)

= 1.96m/s^2 (A moving up, B moving down)

Now find out how fast (velocity) particle A & B are moving when B hits the floor after falling the distance of 0.25m:

d=1/2*a*t^2 and v=a*t => t=v/a, substitute into equation for d:

d=1/2*a*(v/a)^2 = 1/2*a*v^2/a^2 = 1/2*v^2/a

solve for v:

v= (2*a*d)^0.5 = (2*(1.96m/s^2)*(0.25m))^0.5 = 0.99m/s

Now, when B hits the floor it (and A) will have been travelling at 0.99m/s. A will continue to move up at 0.99m/s and the string will go slack. Then gravity will slow it down at the rate of 9.8m/s every second until it stops and falls back down to its original position.

This time is given by 2*v/g (twice the time it takes gravity to stop the vertical motion of A). When A returns to the same position it was in when B hit the floor the string will tighten and jerk B up again, so:

t=2*v/g = 2*(0.99m/s)/(9.8m/s^2) = 0.20 seconds

I hope this helps

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