Calculating Friction Force on a Resting Ladder: A Physics Problem

[TyErd]

Homework Statement

The uniform 30kg ladder rests on the rough floor for which the coefficient of static friction is 0.9 and against the smooth wall at B.

Determine the horizontal force P the man must exert on the ladder in order to cause it to move.

Homework Equations

F= friction coefficient* Normal force

equilibrium equations for horizontal and vertical components and moments (torque).

The Attempt at a Solution

I have attempted the solution but I have no answer to compare with so I am not sure if I am correct or not.

 

[TyErd]

anyone...

 

[TyErd]

i got an answer of 310 is that right?

 

[TyErd]

is the question too difficult to answer or...

 

[xiaoB]

If no going wrong, my answer is 11.25 N .

But next time better post your solution because no guys will calculate for you.

Thanks!

 

[TyErd]

i think that's wrong, has to be higher because the weight force alone is 300N.

 

[TyErd]

what i did was made horizontal components equal zero vertical components equal zero and then moment about B equal zero. but i still don't get an appropriate answer.

 

[xiaoB]

emh...because you just want to to find out the the force horizontal acting on the point B and that is the force for P,so the answer what i get is 112.5 N before is wrong.

I am also no sure for it, i just think about like that and get the answer.:shy:

 

[xiaoB]

But if you know the answer why don't try to show up the work and let the people to check about it because normally no people want to to calculate for you...

 

[TyErd]

y components: N_a - 300 = 0, thus N_a = 300N
x components: N_b + p - F_a = 0
N_b + p - (0.9*300) = 0

moment at B: 300(1.8) - 300(0.9) - 270 (2.4) + p(1.2) = 0, which gives an answer of 315 for P. which i don't think is right considering the weight force and friction.

 

[xiaoB]

"

y components: N_a - 300 = 0, thus N_a = 300N
x components: N_b + p - F_a = 0
N_b + p - (0.9*300) = 0

moment at B: 300(1.8) - 300(0.9) - 270 (2.4) + p(1.2) = 0, which gives an answer of 315 for P. which i don't think is right considering the weight force and friction.

"

Can you tell me have you considered the delta,θ ? And when you want find up the force of x,y components, you should apply cosine and sine. :shy:

 

[TyErd]

umm do you have to? is there a vertical component of P?

 

[xiaoB]

I have to notice that what i have done doesn't mean right,just want to tell how i sum of moment of B in direction x .

After my solution you better check it by yourself because i am not sure what i applied at component x mean cos and sin.

The angle is 53.13^o

(1)The force of P is plus the force acting on the point B equal to the Friction force.

(2)Therefore,to find up the force acting lie on the wall applied sum of moment of B of direction x = 0:

-3xsin53.13xF_B+1.5x300xcos53.13^o=0

F_B=112.5N

(3)F_p=F_f-F_B

ans: F_p=270-112.5=157.5 N

Sorry ! i have changed it so much times because i have checked it again and again.Actually, you are right but just not think above the angle.

Very sorry for you!

 

[ghadz8]

tyerd, are you doing eng1040 too :P