physics buffs?

This might be a longshot but you never know. I need help solving a problem. I just dont know why i cant get it. The answer is 246N.

The q.

A block of mass 20 kg is at rest on a horizontal surface. A force F = 100 N is applied as shown in the drawing, where the angle that the force makes with the horizontal is 30°. The coefficient of static friction between the block and the floor is μ = 0.45 . What is the magnitude of the normal force acting on the block?

 

The second part asks What is the magnitude of the force of friction acting on the block? which is easy. Just (100cos30)/.45 which is just 87N. So where am I going wrong with the first part?

 

You have two forces acting on the block. The first force (the one you can NEVER forget) is gravity. I do believe this will account for most of the force. You also must break up the applied force into its horizontal and vertical vectors. This is done with trig. Let me know if that doesn't make sense. Basically, the horizontal vector will have the majority of that 100N force, but there will still be some in the vertical (into the table).

 

Correct, the normal force is 246.2 N.

There's two forces acting on the block which contribute: gravity and the 100N force.

sin(30)*100=50N

m*g=20*9.81=196.2N

50+196.2=246.2N


To continue, this makes the frictional force:

F_N*mu=246.2*.45=110.79N

This is the force that must be overcome in order to move the block. The horizontal component of the 100N force might be in effort to overcome this force, however:

cos(30)*100=86.6N

meaning the block will not move as 86.6<110.79.

 

wow, this makes me remember why i had a problem with high school, it was grammer school wollllllllll . James

 

A solid sphere of uniform density starts from rest and rolls without slipping a distance of d = 5 m down a q = 23° incline. The sphere has a mass M = 6 kg and a radius R = 0.28 m.

 

Of the total kinetic energy of the sphere, what fraction is translational?

KEtran/KEtotal =


What is the translational kinetic energy of the sphere when it reaches the bottom of the incline?
KEtran =


What is the translational speed of the sphere as it reaches the bottom of the ramp?
v =

4)
Now let's change the problem a little.
Suppose now that there is no frictional force between the sphere and the incline. Now, what is the translational kinetic energy of the sphere at the bottom of the incline?

KEtran =

 

I know this isn't a physics forum but I think reefers can do It better. Any help? I'm lost

 

Bump. Evolved, I know you can do this one