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Friction Example 6 - Friction - Engineering Mechanics

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•
April 9, 2022
by
Ekeeda
YouTube video player
Friction Example 6 - Friction - Engineering Mechanics

TL;DR

Calculate the force needed to prevent a block from sliding down an inclined plane.

Transcript

hello students let us take some problems on the topic of friction i am marking question number one we will read this question first for the arrangement as shown in figure calculate value of force p to be applied in order to avoid the block from sliding down take mu is equal to 0.58 here the diagram has been provided so this is a question of frictio... Read More

Key Insights

  • 🏋️ The force required to prevent a block from sliding down an inclined plane depends on the inclination, weight of the block, coefficient of friction, and the perpendicular component of weight.
  • 🛝 The parallel component of weight contributes to the block sliding down, while friction stops the block from sliding.

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Questions & Answers

Q: What is the significance of preventing the block from sliding down the inclined plane?

The block sliding down would indicate that there is not enough force to counteract the component of weight acting parallel to the plane, leading to motion in the downward direction.

Q: How do we calculate the parallel and perpendicular components of weight?

The parallel component is given by 600 * sine(40 degrees), while the perpendicular component is given by 600 * cosine(40 degrees).

Q: What is the formula for frictional force?

The frictional force (F) is equal to the coefficient of friction (mu) multiplied by the normal reaction (rn).

Q: How can we calculate the value of force P?

Using the equilibrium conditions, we can set up an equation - P = 600 * sine(40 degrees) - F, where F = mu * rn. By substituting the value of rn, we can calculate the force P.

Summary & Key Takeaways

  • The question involves calculating the force (P) required to prevent a block from sliding down an inclined plane with an inclination of 40 degrees.

  • The weight of the block is 600 Newtons, and it has two components - one parallel to the plane and one perpendicular.

  • Using the conditions of equilibrium, we can solve for force P by considering the sum of forces parallel and perpendicular to the plane.


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