Can a 100kg object on a rough surface be moved with a 30° angle force of 800N?

[hastings]

Homework Statement

we want to move an object with mass m=100kg which is still on a rough surface with a force F=800N. between the surface and the object there's a $$\mu_s=1$$. Prove that the object can be moved if the force is applied with a 30° angle with the surface.

Homework Equations

F=m*a (N's 2nd law)

The Attempt at a Solution

friction: A=-mu *N=-100*10=-1000N
F_rx=m*a_x ==> a_x=(-mu*N + F*cos30°)/m=-3.07 m/(s*s)

F_ry=m*a_y ==> a_y=(F*sin30)/m=4

F_r=sqrt( F_rx^2 +F_ry^2)

But I get that A>F_r (result of all forces) so the body doesn't move.

Where am I wrong?

 

[hastings]

someone please help me!

 

[Niishi]

Hello.I can't follow your answer what is N? But friction is Mu times the reaction which is mg - sin 30. F cos 30 is in fact greater than mu times mg- sin 30.Therefore object moves.Hope it is of help.

 

[hastings]

N is the normal force to the surface, what you called reaction. sorry didn't get your suggestion: why is Normal force (N)

$$N= mg - \sin 30$$ ?

I thought N was canceled by $$\vec P= - m \cdot \vec g$$

I know that F has two components, x and y

$$F_x= |\vec F| \cdot \cos 30$$

$$F_y= |\vec F| \cdot \sin 30$$

but N has only one component: along the positive y

 

[hastings]

This is what I figure out of the forces applied to the object. (see the attachment)

 

[Doc Al]

You must solve for the normal force by considering the vertical components of all forces acting on the object: They must sum to zero. In your diagram it looks as if you just assume the normal force equals mg, which is not true in this case, since other forces act vertically.

 

[hastings]

I spoke to my teacher and I understood my mistake.
I was thinking that the reaction force to the surface was ALWAYS only "m∙g". Actually the normal force varies, it's not always "m∙g" or, better, it's not just that.
The gravity force balances the normal force AND the F_y=F∙sin30°

mg= N+ F∙sin30° → N= mg - F∙sin30°.

Once found N, I can calculate the friction force A=µ∙N.
I must then check whether it is A<F∙cos(30°). If so the body moves :)

 

[Niishi]

The normal reaction force of the surfce on the object is equal and opposite to the force exerted by the object on the surface which in this case is not mg but mg - F sin 30.

 

[hastings]

so you are saying N=mg - F∙sin(30°) is not right?

 

[Niishi]

No. I am saying its right ... but u said that the normal force varies i wanted to add that it is equal n opposite to the force that the object exerts on the surface. So u knew that i think?

 

[hastings]

ok, thank you very much!