What is the magnitude of the frictional force acting on the crate?

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Homework Help Overview

The problem involves a crate with a mass of 50 kg being pushed across a level floor by a horizontal force of 150 N, resulting in a constant acceleration of 1.2 m/s². The objective is to determine the magnitude of the frictional force acting on the crate.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law (F=ma) and the forces acting on the crate, including the applied force and friction. Some express confusion about incorporating acceleration into their calculations, while others suggest using a free body diagram to analyze the forces.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the forces involved. Some guidance has been offered regarding the use of free body diagrams and the relationship between the applied force and friction, but no consensus has been reached on the correct approach to find the frictional force.

Contextual Notes

Participants note the importance of considering both the applied force and the frictional force in their calculations, as well as the need to account for the crate's acceleration. There is an acknowledgment of potential misunderstandings regarding the formulas used.

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Homework Statement



A crate with a total mass of 50 kg is initially at rest on a level floor. A horizontal force of 150N pushes the crate across the floor at a constant acceleration of 1.2m/s2. What is the magnitude of the frictional force acting on the crate?

Homework Equations





The Attempt at a Solution


So this question seems so basic but I know I'm doing it wrong. All i did was plugged the numbers into the F=[tex]\mu[/tex]N formulas and got [tex]\mu[/tex]= .31. But in using that formula I am not even taking into account the acceleration and I don't know what to use in order to take it into account?! Someone please explain
 
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It's just F=ma on the crate, but remembering that there are 2 forces acting here; the 150N and the friction.
 
forces.gif
 
ThyQuotidian said:
forces.gif

It means that if 50N force is foreword,then frictional force must act backward.
let net force be fn.
so according to the free body diagram,
fn=150-f (say frictional force be f) ...1

fn=ma=50*1.2
Now from this you'll get fn & from equation 1, you'll get f.
 

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