How Much Can a Fireman Weigh to Safely Climb a Ladder Leaning at 45 Degrees?

[Rha1828]

A uniform ladder of a length 5m and a mass 15 kg leans against a wall, making a 45 degree angle with the ground. The wall is capable of supporting a maximum of 500 N.

A fireman must climb the ladder to it's midpoint, what is the max mass that the fireman can be without collapsing the wall.Ok...So I know that for a uniform problem we know that this means that the Center of Mass will be halfway along the lengh of the board.

Again since this is a center of mass problem the sum of the torques must equal zero.

Et=0

I assumed that I would need to take the (max force)(cos 45) + (Weight of ladder)(distance of ladder) then divide that by the (total distance of the ladder)The answer should be 86.95kg

But I got..

(500N)cos45 + (15kg)(5m)

= 353.33N + 75 kg.m
= 428.33 N.kg.m / 5.0m = 85.66 = 86 kg

I am doing something wrong. I am not able to cancel out some of my units. Its seems like a simple mistake but any advice would help!

Thanks!

 

[sArGe99]

Force X Perpendicular distance = Torque.
Net torque = 0
500 X 5 cos45 = (15+m) g X 2.5 cos 45
m = (1000/9.8) - 15 = 87 Kg

 

[tiny-tim]

Welcome to PF!

Hi Rha1828! Welcome to PF!

A uniform ladder of a length 5m and a mass 15 kg leans against a wall, making a 45 degree angle with the ground. The wall is capable of supporting a maximum of 500 N.

A fireman must climb the ladder to it's midpoint, what is the max mass that the fireman can be without collapsing the wall.
…
I assumed that I would need to take the (max force)(cos 45) + (Weight of ladder)(distance of ladder) then divide that by the (total distance of the ladder)

I don't understand this at all …

you should be taking moments (torques) about the base of the ladder …

that's ∑(force x perpendicular distance)

try again

sArGe99, on this forum, please don't try to give complete answers​