What is the tension in each cable holding up a 100kg Halloween decoration?

AI Thread Summary
The problem involves calculating the tension in three cables supporting a 100kg Halloween decoration, each making a 30-degree angle with the building. The vertical component of the tension is expressed as Tcos30, leading to the equation 3Tcos30 = 100(9.81) for equilibrium. After some calculation errors were acknowledged, the correct tension was determined to be approximately 283.2N. The length of the cables is likely included to emphasize the symmetry of the setup. The discussion concludes with a focus on ensuring accurate calculations and understanding the problem's symmetry.
hamzaarfeen
Messages
4
Reaction score
0

Homework Statement


A huge Halloween decoration of mass 100kg is suspended by 3 cables. Each cable is exactly 25m long and is attached to each of the three building equidistant from each other. Each cable makes an angle of 30 degrees with the side of the building to which it is attached. What is the magnitude of the tension in each cable?

Homework Equations


fnet=ma

The Attempt at a Solution


vertical component for all cables= Tcos30
fnet in y direction = 0
fnet=0
3(Tcos30)=Fg
3Tcos30=100(9.81)
T= 283.2N
 
Physics news on Phys.org
Looks reasonable apart from the numerical value you got in the end, what is your question? Did you multiply the RHS with cos30 instead of dividing when solving for T?
 
Orodruin said:
Looks reasonable apart from the numerical value you got in the end, what is your question? Did you multiply the RHS with cos30 instead of dividing when solving for T?

oh you sorry i made a mistake in calculations...but i still don't know if my answer is right. and what is the purpose for giving the length of the cable?
 
They probably just wanted to make really really sure that you treated the problem as symmetric under 120 degree rotations.

So what do you get now?
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Finding the tension of two cables holding an object at rest
Replies
7
Views
2K
Tension and net torque in a cable
Replies
10
Views
5K
What is the tension in the lift cable and in the string?
Replies
19
Views
3K
What Are the Tensions in Each Cable of a Suspension Bridge?
Replies
31
Views
4K
What are the tensions on two cables holding a girder hanging from a crane?
Replies
14
Views
3K
What is the Tension in Each Cable Supporting the Lamp?
Replies
6
Views
10K
How do I find the tension in two cables holding up a sign at a 50 degree angle?
Replies
5
Views
2K
Tension and Equilibrium: Hanging sign
Replies
8
Views
3K
Cable car system : Tension in the pull cable
Replies
4
Views
7K
Static Equilibrium and tension of cables
Replies
7
Views
3K

Hot Threads

Recent Insights

Back
Top