what is the angle θ which the left-hand cable makes with respect to the wall?

You lot are using an out of date browser. Information technology may not brandish this or other websites correctly.
You should upgrade or use an culling browser.

• Forums
• Homework Aid
• Introductory Physics Homework Aid

Need to find the tension at T in left-hand cable

• Thread starter bikerkid
• Start date Oct six, 2007

Hi. I'm new here and in demand of some help.

I have a problem here in which two cables are supporting a weight of 631 N. The correct-hand cable has a tension of 680 N and and makes an angle of 32 degrees with the ceiling. The left-hand cablevision has a tension of T and makes an angle of X at the wall.

I need to find the tension at T in left-mitt cable in respect to the wall.

I besides need to find the angle that the left-manus cable makes with respect to the wall. I thought it should be 58 degrees, but apparently that is not correct.

Answers and Replies

Hi bikerkid. try to write out these equations:

[tex]\Sigma{F_x} = ma_x[/tex] which becomes [tex]\Sigma{F_x} = 0[/tex]

and [tex]\Sigma{F_y} = ma_y[/tex] which becomes [tex]\Sigma{F_y}=0[/tex].

And so if T is the tension of the unknown then the vertical component is Tsin(X). Horizontal component is Tcos(X).

Cheers. I got the outset part correct. I just don't know what to do for the second part.

Thank you. I got the offset role right. I but don't know what to do for the 2d role.

cool. tin can you bear witness how you got the first part? that way it'll exist easier to bear witness what y'all need to exercise side by side.

I found the horizontal, vertical, and hypotenuse of the known side. And then, I knew that the unknown horizontal was equivalent to the known horizontal. I knew the forces had to add upwards to aught, due to it being in equilibrium, so I solved for the unknown tension and plant it to be 637.029 N. I solved for the other unknown perpendicular but do so and found that to be 270.6557549 N. I thought this angle would be arc tan, but that is not working. I don't know what to do now.

I institute the horizontal, vertical, and hypotenuse of the known side. And then, I knew that the unknown horizontal was equivalent to the known horizontal. I knew the forces had to add up to cipher, due to it existence in equilibrium, so I solved for the unknown tension and found information technology to be 637.029 Due north. I solved for the other unknown perpendicular simply do so and found that to be 270.6557549 N. I thought this angle would be arc tan, but that is not working. I don't know what to do now.

it's arctan(reverse/adjacent)... or arcsin(reverse/hypotenesue) = arcsin(270.65575/637.029) = 25.1 degrees. Is this what you got?

Yeah, that'southward what I got, simply it says the reply is incorrect, even though it says that the force I got was right. Information technology makes no sense.

Yes, that's what I got, but it says the answer is incorrect, fifty-fifty though it says that the force I got was correct. Information technology makes no sense.

Oh... information technology asks for the angle with the wall... I idea the second cablevision was connected to the ceiling too... I think and so the answer is supposed to be 90-25.1 = 64.9 degrees

Hooray. I was going to try that days ago to see if that could be right, but didn't want to bother wasting 1 of my guesses on something that probably wouldn't be correct. Cheers for your aid. I'll probably demand some more assistance tomorrow. Haha.

At that place is some other problem that I don't really understand.

An 8kg block is released from rest on an inclined plane and moves 2.2m during the side by side 3.4s. The acceleration of gravity is ix.8m/s^two.

What is the coefficient of kinetic friction for the incline?

If someone tin can help, that would be really nice. I need the answer to this question one-time before 12:00 tonight (it'due south eight:00 now) to become credit for information technology.

There is another problem that I don't really empathise.

An 8kg block is released from rest on an inclined aeroplane and moves 2.2m during the side by side 3.4s. The dispatch of gravity is nine.8m/due south^2.

What is the coefficient of kinetic friction for the incline?

Did you find the dispatch? What are the forces acting on the cake along the aeroplane?

a = .3806228374 m/due south^2

There are no other forces acting on it.

a = .3806228374 m/s^2

There are no other forces acting on information technology.

aye, that's the correct acceleration.

There are 2 forces... what are they? 1 is friction... what's the other?

Weight? The block has a mass of 8kg. It doesn't say any other forces.

Weight? The block has a mass of 8kg. It doesn't say whatever other forces.

Yeah... divide the weight into 2 components... what is the component forth the plane... what is the component perpendicular to the plane...

Get the equations:

[tex]\Sigma{F_x} = ma_x[/tex]

[tex]\Sigma{F_y} = ma_y[/tex]

you already got [tex]a_x = .3806228374[/tex]. and ay = 0.

72.59N in the Y and 29.3692N in the X ?

72.59N in the Y and 29.3692N in the X ?

What is exactly 72.58 and what exactly is 29.3692N? can you write the equations out...

Westward = mg
Due west = 8(9.8)
W = 78.four

I did F perpendicular as 78.4cos22 and found that to be 72.6912.

I then did 78.4sin22 and found that to be 29.3692.

Alright, I figured information technology out using ii formulas I found online. Thanks for your help.

Alright, I figured information technology out using 2 formulas I found online. Thanks for your assistance.

cool. good job!

Related Threads on Need to find the tension at T in left-mitt cablevision

• Final Mail service
• Oct 3, 2005

• Concluding Mail service
• Mar five, 2013

• Last Post
• Nov 25, 2008

• Last Post
• Nov 7, 2009

• Last Post
• Aug 7, 2017

• Terminal Post
• Aug xvi, 2005

• Last Post
• Jul 18, 2012

• Last Post
• Dec 16, 2009

• Final Postal service
• Oct vi, 2005

• Concluding Post
• Sep 13, 2008

• Forums
• Homework Help
• Introductory Physics Homework Help

cruzmonsuldn1960.blogspot.com

Source: https://www.physicsforums.com/threads/need-to-find-the-tension-at-t-in-left-hand-cable.189423/

Share this post