Bow & arrow -- Forces on the Bow

[Bolter]

Homework Statement

Determine what the magnitude and direction of the force exerted by the archer's left hand

Relevant Equations

SUVAT equations

Hello everyone!

So I have been doing this projectile motion question and do not understand how to do work out which forces act on the archer's left hand in part a)i?

I am specifically told to ignore the weight of the bow & arrow. But this is force that only acts downward for the bow & arrow only, not on the left hand. I can't really deduce what other forces are acting on the left hand in this situation.

Is there a normal force acting on the left hand as a result of supporting weight of the bow & arrow? Since if the archer is initially stationary before firing, then resultant force has to be zero and the body is said to be in equilibrium. I.e. an upward normal force exists on the left hand and is equal in magnitude to the weight of both `& arrow. However I'm not sure if the entire bow & arrow weight is balanced by the normal force?

I have tried the rest but unsure if my answers are right

Here they are

Any help would be appreciated! Thanks

 

[phinds]

I must be missing something. Seems to me this is WAY overly complicated. You have a single system of bow and string, joined as one unit. You have a force in one direction on the string and a force in the other direction on the bow. If the forces are different, wouldn't you expect the bow/string unit to move to the left or right? Since they don't, what does that tell you?

 

[jackwhirl]

Is there a normal force acting on the left hand as a result of supporting weight of the bow & arrow? Since if the archer is initially stationary before firing, then resultant force has to be zero and the body is said to be in equilibrium. I.e. an upward normal force exists on the left hand and is equal in magnitude to the weight of both `& arrow. However I'm not sure if the entire bow & arrow weight is balanced by the normal force?

This is all correct as far as I can tell at a glance, but like phinds said, you're overthinking it. It says to ignore the weight, so ignore the weight. What other force(s) are acting on the bow/arrow/bowstring? That's all you need to consider.

 

[Bolter]

I must be missing something. Seems to me this is WAY overly complicated. You have a single system of bow and string, joined as one unit. You have a force in one direction on the string and a force in the other direction on the bow. If the forces are different, wouldn't you expect the bow/string unit to move to the left or right? Since they don't, what does that tell you?

Both left force is equal to the total right force if the object doesn't move in the horizontal direction.

Well the 220N force that is applied to the left gets balanced by the 2 equal horizontal component of tension. But does this tell me anything about what force is exerted by the left hand?

 

[Bolter]

This is all correct as far as I can tell at a glance, but like phinds said, you're overthinking it. It says to ignore the weight, so ignore the weight. What other force(s) are acting on the bow/arrow/bowstring? That's all you need to consider.

Tension on the string is the only other force that I can think of that is acting on here

 

[jackwhirl]

You're really close. Now, follow that force up the string. Where does it come from? Where does it go?

 

[phinds]

Both left force is equal to the total right force if the object doesn't move in the horizontal direction.

Well the 220N force that is applied to the left gets balanced by the 2 equal horizontal component of tension. But does this tell me anything about what force is exerted by the left hand?

You continue to make this ridiculously harder than it is. Forget everything but the left and right hands.

 

[Bolter]

You're really close. Now, follow that force up the string. Where does it come from? Where does it go?

I suppose I need to consider the vertical component of tension perhaps

The tension comes from the archer's right hand when the archer pulls the string taut backwards? And horizontal component of tension is exerted on the arrow. Not so sure where the vertical component of tension is exerted on?

 

[phinds]

I suppose I need to consider the vertical component of tension perhaps

The tension comes from the archer's right hand when the archer pulls the string taut backwards? And horizontal component of tension is exerted on the arrow. Not so sure where the vertical component of tension is exerted on?

You continue to make this ridiculously harder than it is. Forget everything but the left and right hands.

And re-read post #2

 

[jackwhirl]

I suppose I need to consider the vertical component of tension perhaps

The tension comes from the archer's right hand when the archer pulls the string taut backwards? And horizontal component of tension is exerted on the arrow. Not so sure where the vertical component of tension is exerted on?

Nothing is exerted on the arrow until it is loosed, other than the minimal force to hold it up against gravity, which you've been explicitly instructed to ignore.

When you pull on a string, it doesn't magically resist that force. Tension is transferred through a taut string to the other end, and whatever it is anchored to. What is this string connected to?

 

[Bolter]

You continue to make this ridiculously harder than it is. Forget everything but the left and right hands.

Sorry this proble

Nothing is exerted on the arrow until it is loosed, other than the minimal force to hold it up against gravity, which you've been explicitly instructed to ignore.

When you pull on a string, it doesn't magically resist that force. Tension is transferred through a taut string to the other end, and whatever it is anchored to. What is this string connected to?

It is connected to the ends of the bow, so tension is transferred to the bow then

 

[jackwhirl]

It is connected to the ends of the bow, so tension is transferred to the bow then

And the bow is not moving, right? Thus the net forces are?

 

[Bolter]

You continue to make this ridiculously harder than it is. Forget everything but the left and right hands.

And re-read post #2

Ok so right hand has a 220N acting leftwards so left hand has a 220N force acting rightwards? I am sorry if this statement sounds awfully wrong

 

[Bolter]

And the bow is not moving, right? Thus the net forces are?

Net force has to be zero

 

[jackwhirl]

Ok so right hand has a 220N acting leftwards so left hand has a 220N force acting rightwards? I am sorry if this statement sounds awfully wrong

Net force has to be zero

You've got it. Put it all together now.

 

[Bolter]

You've got it. Put it all together now.

For the entire system to stay in equilibrium, i.e. the resultant force has to stay zero. A 220N force is exerted by the left hand in the right direction to oppose the 220N force that is exerted by the right hand in the left direction

 

[phinds]

For the entire system to stay in equilibrium, i.e. the resultant force has to stay zero. A 220N force is exerted by the left hand in the right direction to oppose the 220N force that is exerted by the right hand in the left direction

Yes. That's what I told you in post #2. Glad you finally got it. We tend to give pretty decent hints here on PF. You should learn to pay attention to them.

 

[Bolter]

Yes. That's what I told you in post #2. Glad you finally got it. We tend to give pretty decent hints here on PF. You should learn to pay attention to them.

Yes my bad, I really did overthink it

 

[jackwhirl]

You did fine. I bet it was a)ii that threw you off. It broke the bow/bowstring into different components in your mind. The answer to a)i is only obvious if you can treat them as one unit.

I still have to stop and think about how tension works for problems like this.

 

[Bolter]

You did fine. I bet it was a)ii that threw you off. It broke the bow/bowstring into different components in your mind. The answer to a)i is only obvious if you can treat them as one unit.

I still have to stop and think about how tension works for problems like this.

You're certain right it did indeed. I have a habit of always breaking down any force I get into its x and y parts. But this was a good example that taught me you don't necessarily have to think of it like that always, and that it's far more easier to treat it as one whole system.

 

[phinds]

You're certain right it did indeed. I have a habit of always breaking down any force I get into its x and y parts. But this was a good example that taught me you don't necessarily have to think of it like that always, and that it's far more easier to treat it as one whole system.

Right. The first thing you should always do is look at the system as a whole and that will often lead you to the proper way to do whatever breakdown of individual components might be needed (or in this case, not needed).