> #mechanics , #physics > Leaver set up load 40N positioned 0.5m from fulcrum. Effort applied 100N. Calculate the velocity ratio of system if load move to new positions that is twice far from fulcrum while effort distance con?
N
Nitha Mulungu
Aug 18

Leaver set up load 40N positioned 0.5m from fulcrum. Effort applied 100N. Calculate the velocity ratio of system if load move to new positions that is twice far from fulcrum while effort distance con?

1 view
1 view

2 answers

Sort By
C
ChatGPT
Aug 18
To calculate the velocity ratio of the system, we can use the formula:

Velocity Ratio = Load distance / Effort distance

Given:
Load = 40N
Load distance = 0.5m
Effort = 100N
Effort distance = x (unknown)

First, let's calculate the effort distance when the load moves to a new position that is twice as far from the fulcrum:

Load distance (new) = 2 * 0.5m = 1m

Now, we can calculate the effort distance using the principle of moments:

Load * Load distance = Effort * Effort distance

40N * 0.5m = 100N * x
20 = 100x
x = 0.2m

Now that we have the effort distance, we can calculate the new velocity ratio:

Velocity Ratio (new) = Load distance (new) / Effort distance
Velocity Ratio (new) = 1m / 0.2m
Velocity Ratio (new) = 5

Therefore, the velocity ratio of the system when the load moves to a new position that is twice as far from the fulcrum while the effort distance remains constant is 5.
0
G
Google Gemini AI
Aug 18
Initial Conditions:

- Load (W) = 40 N
- Distance from fulcrum (d) = 0.5 m
- Effort (P) = 100 N
- Distance from fulcrum (a) = unknown

Velocity Ratio (VR) = Input Distance (a) / Output Distance (d)

Initial Velocity Ratio:

VR = a / d = a / 0.5

New Conditions:

- New distance from fulcrum for load (d') = 2d = 1 m

New Velocity Ratio:

VR' = a / d' = a / 1

Given:

Effort distance remains constant

Equation:

VR / VR' = d' / d

Substituting Values:

0.5a / a = 1 / 0.5
0.5 = 2

Therefore, the new velocity ratio is twice the initial velocity ratio:

VR' = 2 - VR = 2 - (0.5a / 0.5) = 2a
0

Similar Questions

×
Anonymous





© 2024 - Quanswer