How can I draw free diagram mass on the wall mass

[vrdm]

Homework Statement

I HAVE PROBLEM and please help me
the question is : how can I draw free diagram mass on the wall
mass : 12kg fa : 45 angle 30 Kinetic coefficient of friction:0.25
find net f

Homework Equations

l can not draw the free diagram
fx=0 fy=ma
l don't sure l used this equation correctly

The Attempt at a Solution

yes l tried to solve but my answer is wrong
because the right answer is 80 down

 

[tiny-tim]

welcome to pf!

hi vrdm! welcome to pf!

sorry, but i can't understand what this question is about …

can you give us some more details? ​

 

[tiny-tim]

hi vrdm!

the details we need are:

what is "fa"?

are 45 and 30 angles? if they are, between what lines are they the angles?

what is the 12kg? is it a ladder falling down the wall?

is the wall vertical?

 

[tiny-tim]

hi vrdm! happy new year!

thanks for your pm …

A man holds in his hand a piece of wood in the form of cubic mass of 12 kilograms and Put it on the wallPushing force is 45 N AND IT makes an angle of 30º to the vertical if the coefficient of kinetic between the surface of the piece and the wall 0:25 find net force Affecting in a piece of wood?

fa :it means Pushing force is 45 N AND IT makes an angle of 30º to the vertical
12kg: mass of wood

the question is find Net force​

ah, now i understand!

(except, is the man pushing up at 30° to the vertical, or is he pushing down?)

ok, since the block of wood can only move up or down, the net force must be vertical

the net force will be the sum of three forces: the friction force, the weight of the block, and the vertical component of the 45N …

start by finding the friction force: that's the normal force times µ_k, so first you'll need the normal force, ie the horizontal component of the 45N …

what do you get? ​

 

[vrdm]

thank you

 

[vrdm]

THE QUESTION didn't said the man pushed up or down

 

[tiny-tim]

hi vrdm! thanks for your pm, but in future please reply on the thread (anyone who posts in a thread gets automatic email notification of any future posts) …

hi Mr.tiny-tim
and happy new year
thank you very much for help me
and i tried to answer my question

Fg : 12×9.81 = 117.7
Fk: Fn×0.25
Fn=45sin(30)=
fk=0.25×45sin30=5.6
fay=45cos30=39
39-5.6-117.7=-84
* net f is 84 down

but when l see the answer in the book is 80 down
the answer is right or wrong
THANK YOU ​

your method is correct, except that in your equation 39-5.6-117.7 = -84, both your net force and your friction are downward …

if the block of wood is moving downward, then the friction must oppose that: it must be upward, which would give you 39+5.6-117.7 = -73

unfortunately, that's still not -80

(i've tried it with 60° instead of 30°, and pushing up either up or down, but that doesn't give -80 either)

sorry, but i can't see what's wrong ​

 

[vrdm]

thank you for help me to answer my question
and may be the answer in the book is wrong
thank you too much