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Yuri [45]
3 years ago
14

Calculate the following forces for a box with a mass=20.0 kg sitting on an incline of 29.0 degrees and a coefficient of friction

which is 0.455. Calculate the parallel force, normal force, and frictional force.
Physics
1 answer:
Anna007 [38]3 years ago
3 0
Vertical force on the box=mg 
<span>the component of gravity parallel=mg*SinTheta </span>
<span>the component of gravity normal=mg*CosTheta </span>
<span>frictional force up the plane: mg*cosTheta*mu max, but if it is sitting still, it is equal and opposite to mg*cosTheta (it cannot be greater than this or it would go up the plane).</span>
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What could be the possible answer to the question ?<br><br>thankyou ~​
Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

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Response:

  • The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

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Therefore;

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Which gives;

M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}

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Which gives;

F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2}  \approx  \mathbf{0.377  \cdot M \cdot g}

  • F₀ ≈ <u>0.377·M·g</u>

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Answer:

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I am pretty sure about these answers.Thermal goes in the 4th blank.
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