Determining the Minimum Force to Push a Crate

[semc]

You try to push a crate of weight M with a force F on a horizontal floor. The
coefficient of static friction is μs, and you exert the force F under an angle φ below the
horizontal.
Determine the minimum value of F that will move the crate.

Okay so basically what i did was equate Mμs=Fcos φ and evaluate for F. however the answer my teacher gave me is slightly different. He use Nμs=Fcos φand he found N by evaluating the forces in the y component. So the question is why can't i just equate the force applied with the static friction?

 

[pgardn]

You try to push a crate of weight M with a force F on a horizontal floor. The
coefficient of static friction is μs, and you exert the force F under an angle φ below the
horizontal.
Determine the minimum value of F that will move the crate.

Okay so basically what i did was equate Mμs=Fcos φ and evaluate for F. however the answer my teacher gave me is slightly different. He use Nμs=Fcos φand he found N by evaluating the forces in the y component. So the question is why can't i just equate the force applied with the static friction?

Does your M stand for mass? If so your units will come out wrong for force.
If one pushes (y-component) down on a mass sitting on a surface then the Normal force is going to be different compared to the same mass pushed on perfectly horizontally, or the mass pushed up. So in your problem you are pushing horizontally and vertically.

 

[RoyalCat]

Does your M stand for mass? If so your units will come out wrong for force.
If one pushes (y-component) down on a mass sitting on a surface then the Normal force is going to be different compared to the same mass pushed on perfectly horizontally, or the mass pushed up. So in your problem you are pushing horizontally and vertically.

His M is weight, not mass, so unit-wise it works.

But the problem is just as you've pointed out.

The definition of the kinetic frictional force is: [tex]f_{s max}=\mu_s N[/tex] and NOT: [tex]f_{s max}\neq \mu_s W[/tex]

By pushing up on the mass, you're decreasing the normal force from the ground, thereby decreasing the friction force, and making life easier for yourself.

 

[pgardn]

His M is weight, not mass, so unit-wise it works.

But the problem is just as you've pointed out.

The definition of the kinetic frictional force is: [tex]f_{s max}=\mu_s N[/tex] and NOT: [tex]f_{s max}\neq \mu_s W[/tex]

By pushing up on the mass, you're decreasing the normal force from the ground, thereby decreasing the friction force, and making life easier for yourself.

Oops my bad as they say.

That is a really a potentially confusing symbol to use for weight...