MHB Calculating Force for Lever Push - 600N

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To achieve a brake force of 600N with a friction coefficient of 1.2, the necessary force $F_1$ for the lever must be calculated based on the lever arm lengths of 60 cm and 1.3 m. The initial equation proposed, $$(a+b)F_1>F_2 \cdot 1,2 \cdot a$$, is deemed incorrect. The correct approach involves identifying all external forces and applying static equilibrium conditions, ensuring that the sum of horizontal forces, vertical forces, and moments equals zero. This method will lead to the necessary equations to solve for $F_1$. Proper analysis and calculations are essential for determining the force required to push the lever down effectively.
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Hey! :o

With a simple brake block we want to produce brakeforce of 600N. The friction number is 1,2. Which has to be the $F_1$ so that the lever can be pushed down??
a=60 cm, b=1,3 m

View attachment 4340

Does the following have to stand??

$$(a+b)F_1>F_2 \cdot 1,2 \cdot a$$
 

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Hi! (Wave)

mathmari said:
Does the following have to stand??

$$(a+b)F_1>F_2 \cdot 1,2 \cdot a$$

No. This will not be true. (Worried)

The method to solve a problem like this, is:
  1. Make an inventory of all external forces. Which 3rd force is there? (Wondering)
  2. Apply the static equilibrium conditions:
    1. Sum of the horizontal forces is zero.
    2. Sum of the vertical forces is zero.
    3. Sum of the moments (also known as torques) is zero.
    What are the resulting equations? (Wondering)
  3. Solve the equations.
  4. Repeat for the lever only (without the wheel).
 
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