Giving it a try

[Anonymous No.7530867]

Long god damn day!!!

I've got a physic problem here that bothers me a lot.

We have two block, first one of 2kg and second one of 1kg, touching on a horizontale plane. I apply a push of 6 newton on it at a angle of 30 degree. Kinetic friction of 0.250. What is the force between both block.

I tought i'd go for a whole system of 3kg and check the acceleration from there. But alas. when i do this i get friction force of 6.6 and push force of 5.19... wich means the block is going the other way cause acceleration -0.41... if its going the other way then friction should push the other way?... so fucking lost and tired :p


I know you guys don't answer homework problem i'm just shooting blind here cause i'm stuck in what i consider a easy problem so meh why not :p

[Anonymous No.7530874]

I just drew it. I'll work (again) on finding the answer if i get i'll post :)

[Anonymous No.7530876]

I might send some nude of a friends? :O sorry i rarely come here i don't know what are the customs :P i hang out in different...board

[Anonymous No.7530882]

i'm too lazy to check even for tits, but i think you're suppose to consider them both separatly when come with a common equation for both. Fucker

[Anonymous No.7530888]

Two days straight, maybe more, that this board is being flooded by stupid troll threads, or "do my homework" threads like this. The sticky says explicitly to not open these such threads, but apparently no mod is interested in enforcing the rules.

[Anonymous No.7530890]

Calculate acceleration. The acceleration is the same for both objects, because they are 'linked'. Then, you can determine the force on the small block, F1 = a*m1. That force is the force that the big object exerts on the small one

[Anonymous No.7530894]

force on the second block will be the your initial force horizontal wise (cos) - kinetic friction of first block
you never said to which block the friction applies to though

[Anonymous No.7530902]

If friction applies to both blocks:
The total possible friction force is greater than the horizontal component of your "push" force.
Ergo: no movement

[Anonymous No.7530912]

>>7530888
i was not meant to troll i'm really sorry :(

>>7530890
The acceleration i get from the push to a block of 3kg (looking at it as one big object) is negative... wich doesn't make sense cause that would mean i need to change the sense of the friction
>>7530894
i got that but i wanted to get acceleration first :P and you a right friction on both block and in-between block too
>>7530902
that's what i tought but does that mean there is no force between the two cubes? should i consider a=o and thus make my calculation from there to find the push thez give each other

[Anonymous No.7530914]

>>7530912
Max friction force of left block alone is still greater than the horizontal component of the initial force...

[Anonymous No.7530918]

>>7530914
eum correct me if i'm wrong but
Fp=6cos30
Ff=(2*9.81-6sin30)0.250
Fp-Ff=6cos30-(2*9.81-6sin30)0.250=1.04115242271 wich is not negative ?

[Anonymous No.7530924]

>>7530918
Ff=(2*9.81+6sin(30))*0.25=5.6N
Fx=6cos30=5.2N

[Anonymous No.7530925]

>>7530918
The push from above adds to the total normal force on the floor.

[Anonymous No.7530934]

>>7530924
>(2*9.81+6sin(30))*0.25
doesn't = 5.6

[Anonymous No.7530939]

>>7530934
(19.6+3)*0.25
22.6*0.25
5.6