1 - 20 of 31 Posts

#### legoman786

·
##### Starting to become a BOFH
Joined
·
10,471 Posts
Discussion Starter
I am stuck on a question regarding rotational dynamics.<br />
<br />
One end of a rope is tied to the handle of a horizontally-oriented and uniform door. A force <b>F</b> is applied to the other end of the rope as shown in the drawing (can't show you >.< no cam). The door has a weight of 145N and is hinged on the right. What is the maximum magnitude of <b>F</b> at which the door will remain at rest?<br />
<br />
The force is at 20 degrees above the door and to the right. The length of the door is 3.13m and the length from the hinge to the handle is 2.5m.<br />
<br />
I know the <b>F=ma</b> and <b>T=rFsin(angle)</b>, this is also an equilibrium problem. I just can't figure out how to put the numbers into an equation >.< I know that center of gravity has something to do with it.<br />
<br />
EDIT: I think I should point out that I missed the day he was lecturing on this, so I have no notes whatsoever and I am taking all of this out of his powerpoint presentation.

#### legoman786

·
##### Starting to become a BOFH
Joined
·
10,471 Posts
Discussion Starter
Nobody?

#### dkdeath

·
##### Registered
Joined
·
303 Posts
give me 1-2 min and ill post the answer

·
##### Registered
Joined
·
2,823 Posts
i done this a while back cant rember tho <img src="/images/smilies/frown.gif" border="0" alt="" title="Frown" class="inlineimg" /><br />
<br />
I will have a look though

#### legoman786

·
##### Starting to become a BOFH
Joined
·
10,471 Posts
Discussion Starter
Thanks guys

#### dkdeath

·
##### Registered
Joined
·
303 Posts
The force is at 20 degrees above the door what does this mean ?

#### legoman786

·
##### Starting to become a BOFH
Joined
·
10,471 Posts
Discussion Starter
How come I didnt think of this b4?<br><a href="http://assets.overclock.net.s3.amazonaws.com/4/47/475bbeb5_vbattach34136.jpeg"><img src="http://www.overclock.net/image/id/616339/width/525/height/525/flags/LL"></a>

#### Chaogod87

·
##### Registered
Joined
·
481 Posts
Ok, since the door is on the floor horizontally, u just need enough force the lift the door.<br />
<br />
So, which way does the door move initially,... Up (because tehre are hinges on the right so it cannot love right). So you need to calculate the y (vertical) component of F, which is F*sin(20) = weight (145N). Once the door is lifted the x-component of the force will take care of it because the angle gets bigger, then torque will get bigger. Also the hinges will push up the door.<br />
<br />
I am not sure what the hinges will do to the y component, in initial state.<br />
<br />
Since this is a static problem, the x and y components have to add up, meaning i am assuming the hinges only contribute in x axis (initially) because the door is horizontal.<br />
<br />
Don't take my word for it. But thats how i would think of this problem.<br />
<br />
Good luck

#### apavlov13

·
##### Registered
Joined
·
787 Posts
I was thinking that too, and that would be the way to go if you haven't gone over integrals yet. Because the door is initially horizontal, it will have to travel upwards, so Fnet = Fsin(20) - mg. Fnet = 0, mg = 145N.<br />
<br />
Mkay after a bit of thought, the door is supported by both the rope and the hinge, Fnet = Fsin(20) + (normal force at hinge) - mg.

#### legoman786

·
##### Starting to become a BOFH
Joined
·
10,471 Posts
Discussion Starter
But I'm looking for F. I'm beginning to implode from all the different equations I have tried. All of them require F.

#### apavlov13

·
##### Registered
Joined
·
787 Posts
Theres going to be a certain split of the weight between the hinge and the rope. Lets say the hinge supports 30% of the weight(guess). That means the rope supports 70% of the weight, or 101.5N. F = 101.5/sin(20). F=297N.<br>
That is not the right answer unless the Hinge does support 30% of the weight. That's what you have to figure out.

#### legoman786

·
##### Starting to become a BOFH
Joined
·
10,471 Posts
Discussion Starter
The door's uniformly built, meaning... nvm.<br><br>
Would it help if I said that this is multiple choice?<br>
145N, 265N, 381N, 424N, and 530N.

#### OCZedd

·
##### Registered
Joined
·
2,439 Posts
Isnt it like this (see attached)?<br><br>
Edit: Ok I got it. Its F = 145N/cos70 = 423.95N ~ 424N <img alt="" class="inlineimg" src="/images/smilies/biggrin.gif" style="border:0px solid;" title="Big Grin"><br><a href="http://assets.overclock.net.s3.amazonaws.com/f/f9/f910eaa1_vbattach34137.png"><img alt="LL" src="http://www.overclock.net/image/id/616338/width/525/height/525/flags/LL" style="width:525px;height:290px;"></a>

#### legoman786

·
##### Starting to become a BOFH
Joined
·
10,471 Posts
Discussion Starter
Yea, but how would the cos help?<br><br>
EDIT: I'm sorry for not knowing, like I said, I missed the day he was lecturing on this stuff.

#### apavlov13

·
##### Registered
Joined
·
787 Posts
cos70 = sin20, so its the same. The horizontal component cancels out with the hinge pushing back in the opposite direction. If the rope was connected at the end of the door, then the force would be equally split between the hinge and the rope, but because it isn't I don't know how to do that...yet.<br><br>
Heres what i'm thinkin:<br><br><a href="http://www.overclock.net/attachment.php?s=fb118d83aa0ea05832d812d721028acf&attachmentid=34138" target="_blank" title="">Attachment 34138</a><br>
Fsin20 + F<span style="font-size:small;">(hinge) = 145<br><br></span><br><a href="http://assets.overclock.net.s3.amazonaws.com/2/2e/2e12bc6e_vbattach34138.bmp"><img alt="LL" src="http://www.overclock.net/image/id/616337/width/525/height/525/flags/LL" style="width:525px;height:131px;"></a>

#### OCZedd

·
##### Registered
Joined
·
2,439 Posts
Believe me the answer is 424N (145/sin70)

#### apavlov13

·
##### Registered
Joined
·
787 Posts
Can you explain it then? think about it in real life: I would be easier to lift one end of the door, rather than the whole thing. Lifting the door straight up would be 145N. Lifting it straight up with a force applied 20deg to the door would be 424N. Lifting one end with a rope .625M from the end at 20deg to the door is something else.

#### legoman786

·
##### Starting to become a BOFH
Joined
·
10,471 Posts
Discussion Starter

#### OCZedd

·
##### Registered
Joined
·
2,439 Posts
<div style="margin:20px;margin-top:5px;">
<div class="smallfont" style="margin-bottom:2px;">Quote:</div>
<table border="0" cellpadding="6" cellspacing="0" width="99%"><tr><td class="alt2" style="border:1px inset;">
<div>Originally Posted by <strong>apavlov13</strong> <a href="showthread.php?s=fb118d83aa0ea05832d812d721028acf&p=1382117#post1382117"><img alt="View Post" class="inlineimg" src="http://static.overclock.net//img/forum/go_quote.gif" style="border:0px solid;"></a></div>
<div style="font-style:italic;">Can you explain it then? think about it in real life: I would be easier to lift one end of the door, rather than the whole thing. Lifting the door straight up would be 145N. Lifting it straight up with a force applied 20deg to the door would be 424N. Lifting one end with a rope .625M from the end at 20deg to the door is something else.</div>
</td>
</tr></table></div>
OK, I see where u r coming from. But all the answers he gave are above 145N, this is because of the angle. If the angle was greater, say 50 degrees then it would be 145N/cos(90-50) like before. This comes out to be 189N, its less because more of the force applied from the rope (Tension) is vertical rather than horizontal, as the horizontal component would increase 145N/cos50 = 225N (more is horizontally than vertically. I am saying its more because most of the force is used horizontally rather than vertically. If the angle was striaght up (90 degrees) it would be 145N/cos(90-90) = 145N. <img alt="" class="inlineimg" src="/images/smilies/biggrin.gif" style="border:0px solid;" title="Big Grin"> OK now?

#### apavlov13

·
##### Registered
Joined
·
787 Posts
I got 265 and heres how: I found how much torque is applied by the weight of the door. That means taking the integral form 0 to 3.13 of the radius multiplied by the linear density (145/3.13). I got 226.9N as the torque applied downward. I set that equal to the vertical force applied by the rope. 226.9=rFsin20<br>
F = 226.9/2.5sin20<br>
F = 265.39<br><br><br>
And OCZedd, to finish my argument, I was saying as if the door could only move straight up, like it is on rails, but the force wasn't applied straight up. Think of the donkeys that used to pull barges along canals.

legoman786
1 - 20 of 31 Posts