ne toThe coefficient of static friction between the m = 3.10 kg crate and the 35.0° incline of the figure below is 0.260. What minimum force F must be applied to the crate perpendicular to theprevent the crate from sliding down the incline?35.0°

Name: ne toThe coefficient of static friction between the m = 3.10 kg crate
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Description: ne toThe coefficient of static friction between the m = 3.10 kg crate and the 35.0° incline of the figure below is 0.260. What minimum force F must be applied to the crate perpendicular to theprevent the crate from sliding down the incline?35.0°

Answered: The coefficient of static friction…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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ne to
The coefficient of static friction between the m = 3.10 kg crate and the 35.0° incline of the figure below is 0.260. What minimum force F must be applied to the crate perpendicular to the
prevent the crate from sliding down the incline?
35.0°

Expert Solution

Step 1

Given

The mass of crate is m = 3.10 kg.

The inclined angle is $θ = 35^{o}$ .

The coefficient of static friction is $μ_{s} = 0.260$ .

The free body diagram of the inclined plane is shown as,

For the equilibrium, the force applied on y direction is,

$\begin{array}{rcl} \sum_{F_{y}} & = & 0 \\ N - F - m g \cos 35 & = & 0 \\ N & = & F + m g \cos 35 . . . . . (i) \end{array}$

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