Mechanics issue Watch

OJ Emporium
Badges: 10
Rep:
?
#1
Report Thread starter 1 week ago
#1
"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
1
reply
OJ Emporium
Badges: 10
Rep:
?
#2
Report Thread starter 1 week ago
#2
The picture
Attached files
0
reply
golgiapparatus31
Badges: 11
Rep:
?
#3
Report 1 week ago
#3
(Original post by OJ Emporium)
The picture
Oops! I thought they were rough surfaces!
Last edited by golgiapparatus31; 1 week ago
0
reply
OJ Emporium
Badges: 10
Rep:
?
#4
Report Thread starter 1 week ago
#4
(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?
0
reply
RDKGames
  • Community Assistant
Badges: 20
Rep:
?
#5
Report 1 week ago
#5
(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 mg 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.
Last edited by RDKGames; 1 week ago
1
reply
OJ Emporium
Badges: 10
Rep:
?
#6
Report Thread starter 1 week ago
#6
(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 mg 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?
0
reply
RDKGames
  • Community Assistant
Badges: 20
Rep:
?
#7
Report 1 week ago
#7
(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.
0
reply
OJ Emporium
Badges: 10
Rep:
?
#8
Report Thread starter 1 week ago
#8
Ohh right but hey if it works it works i guess
Thanks man
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of Derby
    Postgraduate and Professional Open Evening - Derby Campus Postgraduate
    Tue, 22 Jan '19
  • University of the West of England, Bristol
    Undergraduate Open Afternoon - Frenchay Campus Undergraduate
    Wed, 23 Jan '19
  • University of East London
    Postgraduate Open Evening Postgraduate
    Wed, 23 Jan '19

Brexit: Given the chance now, would you vote leave or remain?

Remain (1529)
79.3%
Leave (399)
20.7%

Watched Threads

View All