Solving Tension Force in Plank Supported by Cables

AI Thread Summary
To solve the tension force in the cable farthest from the jaguar, consider the plank's static equilibrium and the distribution of forces due to the jaguar's weight. The total tension in the cables must equal the jaguar's weight of 800 N, but since the jaguar is positioned 2.0 m from one end, the tension will not be evenly distributed. By applying the principle of moments around the cable closest to the jaguar, one can calculate the tensions in both cables. The analysis shows that the tension in the cable farthest from the jaguar is determined by balancing the moments and forces acting on the system. This problem exemplifies the principles of statics in physics.
eraemia
Messages
53
Reaction score
0

Homework Statement



A strong plank 8.0 m long and having a negligible mass is supported by steel cables from both ends. A jaguar with a weight of 800 N stands 2.0 m from one end. What is the tension force in the cable farthest from the jaguar?

a. 120 N
b. 200 N
c. 240 N
d. 400 N
e. 600 N
f. 800 N

Homework Equations





The Attempt at a Solution



I'm not sure how to solve this problem. Is it a two dimensional problem? I'm not told whether the bridge bends down because of the jaguar's weight. Now, to find the tension force, I know it's m times a, but since the jaguar is not moving, the jaguar's pull on the rope (the tension force) is only going to be the force of gravity which the jaguar experiences (in the -z direction). Is this then simply 800 N? But how do I take distances into account when calculating tension force? Thanks for any help.
 
Physics news on Phys.org
The problem is one of statics.

The sum of the tensions in the cables must equal that of the jaguars weight, but since the jaguar is asymmetrically positioned between the cables, one will bear more of the weight than the other.

To find the distribution of forces between the cables, pick one end and use the requirment that the sum of the moments about a pivot must equal zero.

Pick the end closest to the jaguar, and determine the moments. Take positive moment to cause counterclockwise rotation looking at the x,y plane.

Reference:
http://web.mit.edu/course/3/3.11/www/modules/statics.pdf
 
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...
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...
Back
Top