Ask Question
6 June, 14:04

A 50.0-kg box is resting on a horizontal floor. a force of 250 n directed at an angle of 32.0° below the horizontal is applied to the box. what is the minimum coefficient of static friction between the box and the surface required for the box to remain stationary?

+2
Answers (1)
  1. 6 June, 16:46
    0
    The minimum value friction is equal to the horizontal force component of 250 N force.

    The horizontal force component can be found by multiplying the 250N force by cosine of 32 degrees. = 212N

    F = μ N

    Hence, divide the horizontal component of the force by the Normal Force.

    Normal Force equals to the Vertical component of the 250N force plus the weight of the box. = 490.5N

    The answer is: 0.432
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A 50.0-kg box is resting on a horizontal floor. a force of 250 n directed at an angle of 32.0° below the horizontal is applied to the box. ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers