Three math instructors are trying to hold a ferocious bear from eating students trapped in a math112 class.?

Question

Three math instructors are trying to hold a ferocious bear from eating students trapped in a math112 class. The bear, in the center , is wearing a collar with 3 ropes attached to it and each instructor has hold of a rope.Scott is pulling in the direction North 60degree West with a force of 350 lb and Nancy is pulling in the direction North 45 degree East with a force of 400lbs. Find the magnitude of force and direction Ken must pull to keep the bear from moving.

Answer 1

Well To solve this problem, you need to break up forces in x and y components. Once you find both x and y components, you can find magnitude of force and direction ken must pull.

To keep the bear from moving, some of the forces (in both x and y direction) equal to zero.

So, sum of the forces in X-direction = 350sin(60) - 400sin(45) - F(in x direction) = 0
Sum of the forces in y-direction = 400cos(45) + 350cos(60) - F(in y direction) = 0

Solve for F(in x direction) and F(in y direction), you will get these: F(in x direction) = 20.266 lb east and F(in y-direction) = 457.84 lb south
To find magnitude = ||F||= sqrt((F(x))^2+(F(y))^2) = 458.29 lb

To find angle= tan(theta) =(F(y))/(F(x)) where theta is the angle
theta = arctan (F(y)/F(x))= 87.5 degree east south

So Ken must pull the rop with force 458.29 lb in the direction East 87.5 degree South to keep the bear from moving.
I think this is right. Good Luck.!!!

Answer 2

dear yth,

no I am not going to be giving you an answer, but just want to point out this purely hypothetical case is really ridiculous in real life, holding the bear with 3 ropes will never able to immobilize the bear. I know the intent of the homework is to find the balance of energy when you pull in 3 directions. But this is silly because the bear can easily upset the balance by simply going in any direction of one of the ropes - this creates a slag and the other 2 person would lose their balance naturally. Thought you might want to know that.

The solution is simple mathematically just draw vector diagrams and calculate them, you might want to use a graph paper to help you though.

Answer 3

Look at this from the plan view (overhead). Draw a diagram of the forces (ropes) you know and their angles, and add the third at an arbitrary angle and magnitude called ø and F. Now sum up the x and y components of all the forces (minding the signs); the x components should sum to zero, as should the y components. You will get two equations for the two unknowns F and ø..