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

AI Thread Summary
To determine the magnitude of the frictional force acting on a 50 kg crate being pushed with a horizontal force of 150 N and accelerating at 1.2 m/s², the net force must be calculated. The net force (fn) is derived from the equation fn = ma, resulting in fn = 60 N. The frictional force (f) can then be found using the equation 150 N - f = 60 N, leading to f = 90 N. This indicates that the frictional force opposing the motion of the crate is 90 N. The solution highlights the importance of considering both the applied force and the resulting acceleration in calculating friction.
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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=\muN formulas and got \mu= .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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