Solving Tension Force in Plank Supported by Cables

[eraemia]

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.

 

[Astronuc]

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

 

[ogb p]

I would approach this problem by first identifying the forces acting on the plank. These would include the weight of the jaguar, the tension forces from the cables, and the normal force from the ground. Since the plank is at rest, we can assume that the sum of all forces in the x and y direction is equal to zero.

Next, I would draw a free body diagram and label all the forces acting on the plank. This will help me visualize the problem and determine which equations to use.

In this case, we can use the equation F=ma to solve for the tension force in the cable farthest from the jaguar. Since the plank is at rest, the acceleration is zero, and we can rearrange the equation to solve for the tension force.

T=mg

Where:
T = tension force
m = mass of the jaguar
g = acceleration due to gravity (9.8 m/s^2)

Plugging in the values given in the problem, we get:

T = (800 N)(9.8 m/s^2)
T = 7840 N

However, this is the total tension force in both cables. Since the jaguar is standing 2.0 m from one end, we can assume that the tension force in that cable is greater than the tension force in the other cable. By using the principle of moments, we can calculate the tension force in the cable farthest from the jaguar.

M1 = M2

Where:
M1 = moment caused by the weight of the jaguar (800 N x 2.0 m)
M2 = moment caused by the tension force in the cable farthest from the jaguar (T x 8.0 m)

Solving for T, we get:

T = (800 N x 2.0 m)/8.0 m
T = 200 N

Therefore, the correct answer is b. 200 N. This is the tension force in the cable farthest from the jaguar.