Two prospectors are pulling on ropes attached around the neck of a donkey that doesn’t want to move. One
prospector pulls with a force of 25 lbs, and the other pulls with a force of 50 pounds. If the angle between the
ropes is 30 degrees, then how much force must the donkey use in order to stay put? (The donkey knows the
proper direction in which to apply his force.)
Since the system is in Equilibrium, the
force of the donkey must be equal and opposite the applied force:
Fd = -(25 + 50[30o]) = -(25 + 43.3+25i) = -(68.3 + 25i) = -68.3 - 25i = 72.7 Lbs[20.1o] S. of W.
To find the force that the donkey must use to stay put, we can use the concept of vector addition.
Let's break down the forces into their x and y components.
The force applied by the first prospector can be split into two components:
Fx1 = 25 lbs * cos(30°)
Fy1 = 25 lbs * sin(30°)
The force applied by the second prospector can also be split into two components:
Fx2 = 50 lbs * cos(30°)
Fy2 = 50 lbs * sin(30°)
Since the donkey wants to stay put, the total force in the x-direction should cancel out. Therefore, the donkey's force in the x-direction will be:
Fx_d = - (Fx1 + Fx2)
The donkey only needs to oppose the forces in the y-direction to stay put. Therefore, the donkey's force in the y-direction will be:
Fy_d = - (Fy1 + Fy2)
Next, we can calculate the magnitude of the donkey's force using Pythagoras' theorem:
F_d = √(Fx_d^2 + Fy_d^2)
Plugging in the values, we get:
F_d = √[(-(25 lbs * cos(30°)) - (50 lbs * cos(30°)))^2 + (-(25 lbs * sin(30°)) - (50 lbs * sin(30°)))^2]
Simplifying this expression will give us the exact value of the force that the donkey must use to stay put.
To find out how much force the donkey must use to stay put, we need to analyze the situation using vector addition.
First, let's draw a diagram to visualize the problem.
^
/
/ 30°
/
50 lbs /-----\ 25 lbs
< >
| |
| Donkey|
| |
| |
\ /
\ /
Now, let's break down the forces into their components.
The 50 lbs force can be broken down into two components: one parallel to the direction the donkey is being pulled (50 * cos(30°)) and one perpendicular to the direction of the pull (50 * sin(30°)).
Similarly, the 25 lbs force can be broken down into two components: one parallel to the direction the donkey is being pulled (25 * cos(30°)) and one perpendicular to the direction of the pull (25 * sin(30°)).
The force exerted by the donkey to stay put must exactly balance the pull from the prospectors. Therefore, the sum of the components parallel to the direction of the pull must be zero.
Let's calculate the components:
Component parallel to the direction of pull for the 50 lbs force:
50 * cos(30°) = 50 * √3/2 ≈ 43.3 lbs
Component parallel to the direction of pull for the 25 lbs force:
25 * cos(30°) = 25 * √3/2 ≈ 21.7 lbs
Since the donkey's force must balance the prospectors' pull, the sum of these components must be zero:
Donkey's force - 43.3 lbs - 21.7 lbs = 0
Solving for Donkey's force:
Donkey's force = 43.3 lbs + 21.7 lbs = 65 lbs
Therefore, the donkey must use a force of 65 lbs in the opposite direction to that of the prospectors' pull in order to stay put.