Diagram for Wooden Block Friction Problem

[takando12]

Homework Statement

A wooden block of mass M resting on a rough horizontal floor is pulled with a force F at an angle θ with the horizontal. If μ is the coeeficient of kinetic friction between the block and the surface, then acceleration of the block is?

Homework Equations

The Attempt at a Solution

[/B]The horizontal component of the force - the friction force = ma. And the formula for kinetic friction is f=μ_kN. The normal reaction will be Mg. But I suppose the vertical component of force must also fit in somehow. Should I add it or subtract it to Mg? my initial hunch was to add it but the answer says it's been subtracted. Can someone explain why?

 

[brainpushups]

The horizontal component of the force - the friction force = ma.

Fine.

And the formula for kinetic friction is f=μkN

Yes

The normal reaction will be Mg

No

But I suppose the vertical component of force must also fit in somehow. Should I add it or subtract it to Mg?

Certainly

my initial hunch was to add it but the answer says it's been subtracted. Can someone explain why?

Have you drawn a force diagram. If you draw the forces with the x and y components it should be clear why they are subtracted. This is what reduces the normal force between the surfaces and hence the force of friction.

 

[feyn_stein17]

I agree. Always draw force diagrams with these types of questions, and be sure to label the components of all the forces acting on the object. Keep in mind any y component of force will affect the normal force on the object, which is why, in this particular case, the normal force in not simply Mg.

 

[haruspex]

Homework Statement

A wooden block of mass M resting on a rough horizontal floor is pulled with a force F at an angle θ with the horizontal. If μ is the coeeficient of kinetic friction between the block and the surface, then acceleration of the block is?

Homework Equations

The Attempt at a Solution

[/B]The horizontal component of the force - the friction force = ma. And the formula for kinetic friction is f=μ_kN. The normal reaction will be Mg. But I suppose the vertical component of force must also fit in somehow. Should I add it or subtract it to Mg? my initial hunch was to add it but the answer says it's been subtracted. Can someone explain why?

It is not stated whether the force F acts at an angle theta above the horizontal or an angle theta below the horizontal, so there is no way to know whether it should be added or subtracted.
If we assume it is above the horizontal, would you add or subtract?

 

[takando12]

I did draw out the force diagram. But I thought I must add them because they are acting in the same direction. I think my basic understanding is still at flaw. I don't understand how the normal reaction is proportional to the frictional force. I suppose the weight is proportional as it would be harder to move a heavy object as it would stick to the surface more. But why the normal reaction in specific?

 

[laplacean]

You can imagine the normal force as "pushing" the two surfaces (the wooden block and the floor) together. It is a rather easy conclusion to arrive to that if there is a lot of force pushing the two surfaces together, then it will be harder to slide one of the surfaces across the other.

 

[haruspex]

I thought I must add them because they are acting in the same direction.

What and what are acting in the same direction, exactly?

I don't understand how the normal reaction is proportional to the frictional force. I suppose the weight is proportional as it would be harder to move a heavy object as it would stick to the surface more. But why the normal reaction in specific?

The surfaces in contact don't "know" anything about gravity. All they care about is the forces tending to move the two surfaces differently.
The harder they are pressed together (i.e., the normal force), the more they can resist being slid against each other (parallel force).

 

[brainpushups]

I did draw out the force diagram

Can you post your diagram?