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The safe allowable tension in a V-belt is 3,8 MPa and its cross-sectional area is 800 mm². What is the maximum tension to which the belt can be subjected?

Physics

ScanSolve AI Answer

Answer

3040 N\text{3040 N}

Step-by-step solution

4 steps
  1. Identify the given values and the relevant formula.

    The safe allowable tension (stress) (σ\sigma) is given as 3.8 MPa3.8 \text{ MPa}.
    The cross-sectional area (AA) is given as 800 mm2800 \text{ mm}^2.
    The relationship between stress, force (tension), and area is:

    σ=FA\sigma = \frac{F}{A}

    where FF is the force or tension.

  2. Convert the stress unit to be consistent with the area unit.

    1 MPa=1Nmm21 \text{ MPa} = 1 \frac{\text{N}}{\text{mm}^2}.
    So, σ=3.8Nmm2\sigma = 3.8 \frac{\text{N}}{\text{mm}^2}.

  3. Rearrange the formula to solve for the maximum tension (FF).

    F=σ×AF = \sigma \times A
  4. Substitute the given values into the formula and calculate the maximum tension.

    F=3.8Nmm2×800 mm2F = 3.8 \frac{\text{N}}{\text{mm}^2} \times 800 \text{ mm}^2 F=3040 NF = 3040 \text{ N}

    The maximum tension to which the belt can be subjected is 3040 N\text{3040 N}.

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