alever is used to lift a heavy load. when a to-n force pushes one end of the lever down 1.2m the load rises 0.2m. show the weight of the load is 300n
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"a lever is used to lift a heavy load. when a 50-n force pushes one end of the lever down 1.2m the load rises 0.2m. show the weight of the load is 300n "
Work done by 50 N force = 50*1.2 =60 N-m
Work done by load = W*0.2
Equate work done
0.2W = 60
W = 60/0.2 N
= 300 N
A lever is used to lift a heavy load. When a 50 N force pushes one end of the lever down 1.2m the load rises .2m. Show the weight of the load is 300N
ADVANTAGE of lever is input/output = 1.2/.2 = 6
W=FD
W=50N*6 (b/c of advantage) = 300N
To solve this problem, we can use the principle of lever and the equation of equilibrium.
Step 1: Recognize the forces acting on the lever.
- There are two forces involved: the force applied to one end of the lever (also known as the effort force or input force) and the weight of the load (also known as the resistance force or output force).
Step 2: Identify the distances involved.
- The distance from the applied force to the pivot point is 1.2 m.
- The distance from the weight of the load to the pivot point is 0.2 m.
Step 3: Determine the relationship between the forces and distances using the equation of equilibrium for levers.
- According to the equation of equilibrium, the effort force multiplied by its distance to the pivot point equals the weight of the load multiplied by its distance to the pivot point.
Mathematically, this can be written as:
Effort force × Effort distance = Load force × Load distance
Step 4: Plug in the given values and solve for the weight of the load.
Let's assume the effort force (force applied) is F and the weight of the load is W. We can substitute the known values into the equation:
F × 1.2 m = W × 0.2 m
Rearranging the equation to solve for W (weight of the load):
W = (F × 1.2 m) / 0.2 m
W = 6F
Given that the effort force is 10N, we can calculate the weight of the load:
W = 6 × 10 N
W = 60 N
Therefore, the weight of the load is 60 N, not 300 N as mentioned in the question.
To solve this problem, we can use the principle of moments, which states that the sum of the clockwise moments is equal to the sum of the anticlockwise moments.
In this case, we have a lever with a force applied to one end and a load on the other. Let's assume that the load is L newtons.
The distance from the applied force to the pivot point is 1.2m, and the distance from the load to the pivot point is 0.2m.
Using the principle of moments, we can get the following equation:
Force × Distance from force to pivot = Load × Distance from load to pivot
This can be expressed as:
To × 1.2m = L × 0.2m
Simplifying the equation, we get:
To = (L × 0.2m) / 1.2m
To = L/6
We are given that the applied force, To, is equal to 10N (from the question). Substituting this value into the equation, we get:
10N = L/6
Multiplying both sides of the equation by 6, we have:
60N = L
Therefore, the weight of the load, L, is 60N.
In conclusion, the weight of the load is 60N, not 300N as mentioned in the question.