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Mechanics issue

"A uniform ladder, AB is leaning across a smooth verical wall on rough horizontal ground at an angle of 70° to the horizontal. The ladder has length 8m and is held in equilibrium by a fractional force of magnitude 60 N acting horizontally at B, as shown in the diagram( basically point A is on the wall and B is on the floor) "

Basically the first question is to find the normal reaction at point A, I'm thinking it's just 60N because f1 = f2 but im unsure if it's even correct

The second is to find the mass of the ladder and I don't understand how I do that, i have equations such as the f1+f2 =mg but i know im supposed to add in the sin/cos signs, but i don't know how
The picture
IMG-20190109-WA0001-1.jpg7.1KB
Original post by OJ Emporium
The picture


Oops! I thought they were rough surfaces!
(edited 5 years ago)
Original post by golgiapparatus31
Take moments about the point vertically below A and at the same horizontal level as B

The frictional force at A will have no moment. Solve for normal contact force at A

Wait so just to be sure I use Fdsin theta at the vertical and Fdcos theta at the horizontal right?
Original post by OJ Emporium
"A uniform ladder, AB is leaning across a smooth verical wall on rough horizontal ground at an angle of 70° to the horizontal. The ladder has length 8m and is held in equilibrium by a fractional force of magnitude 60 N acting horizontally at B, as shown in the diagram( basically point A is on the wall and B is on the floor) "

Basically the first question is to find the normal reaction at point A, I'm thinking it's just 60N because f1 = f2 but im unsure if it's even correct


That's right.

The second is to find the mass of the ladder and I don't understand how I do that, i have equations such as the f1+f2 =mg but i know im supposed to add in the sin/cos signs, but i don't know how


Rather than taking moments at what @golgiapparatus31 said, you should take moments at B. If you take their approach, there is still an issue of not knowing what the reaction force at B is! So taking moments at B will allow you to ignore both the reaction AND friction forces.

That said, it's just simple GCSE pythagoras from there. The weight mgmg is acting downwards at a distance 4m from B up the ladder. So what is the perpendicular distance to this force ?? Note that this distance is horizontal.
(edited 5 years ago)
Original post by RDKGames
That's right.



Rather than taking moments at what @golgiapparatus31 said, you should take moments at B. If you take their approach, there is still an issue of not knowing what the reaction force at B is! So taking moments at B will allow you to ignore both the reaction AND friction forces.

That said, it's just simple GCSE pythagoras from there. The weight mgmg is acting downwards at a distance 4m from B up the ladder. So what is the perpendicular distance to this force ?? Note that this distance is horizontal.

Okay so after a while of going through moments and stuff(now I know when to use sine/cosine), I ended up getting this:

F1 = F2 therefore Force on wall = 60N
Counterclockwise torque = Clockwise torque
Fwall x 8 x sin70 = 4Nsin20 (where N is the reaction force and 4 is half the distance)
480sin70/4sin20 = N = 329.697N

329.697/9.81 = 33.6 kg

Is this right?
Original post by OJ Emporium
Okay so after a while of going through moments and stuff(now I know when to use sine/cosine), I ended up getting this:

F1 = F2 therefore Force on wall = 60N
Counterclockwise torque = Clockwise torque
Fwall x 8 x sin70 = 4Nsin20 (where N is the reaction force and 4 is half the distance)
480sin70/4sin20 = N = 329.697N

329.697/9.81 = 33.6 kg

Is this right?


Yep.

It would've been more obvious if you used cos(70) instead of sin(20) though.
Ohh right but hey if it works it works i guess
Thanks man
Reply 8
Avatar for Aeshakhan
Aeshakhan
9
Original post by OJ Emporium
Okay so after a while of going through moments and stuff(now I know when to use sine/cosine), I ended up getting this:

F1 = F2 therefore Force on wall = 60N
Counterclockwise torque = Clockwise torque
Fwall x 8 x sin70 = 4Nsin20 (where N is the reaction force and 4 is half the distance)
480sin70/4sin20 = N = 329.697N

329.697/9.81 = 33.6 kg

Is this right?


How do you know F1 = F2 ?
Reply 9
Avatar for davros
davros
16
Original post by Aeshakhan
How do you know F1 = F2 ?


assuming one of those is the frictional force and the other is the normal (horizontal) reaction at the wall, then it's just resolving horizontally

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