Friction and torque of ladder resting on wall

AI Thread Summary
The discussion revolves around calculating the torque of a ladder resting against a wall, considering the forces acting on it. The ladder has a uniform mass and angle, with a person standing at a specific distance from the bottom. The key point is that torque is determined by the force multiplied by the perpendicular distance from the pivot point, not just the vertical or horizontal distances. Participants clarify that vertical forces relate to horizontal distances and vice versa for horizontal forces. Understanding this distinction is crucial for solving the problem accurately.
SA32
Messages
31
Reaction score
0

Homework Statement


A uniform ladder with mass m2 and length L rests against a smooth wall. A do-it-yourself enthusiast of mass m1 stands on the ladder a distance d from the bottom (measured along the ladder). The ladder makes an angle x with the ground. There is no friction between the wall and the ladder, but there is a frictional force of magnitude f between the floor and the ladder. N1 is the magnitude of the normal force exerted by the wall on the ladder, and N2 is the magnitude of the normal force exerted by the ground on the ladder. Throughout the problem, consider counterclockwise torques to be positive. None of your answers should involve pi (i.e., simplify your trig functions).

Sum the torques given in the problem about the point at which the ladder touches the wall.

Homework Equations


T(torque)=F*d

See attachment for diagram.

The Attempt at a Solution


I know the answer is,

-m1g(L-d)cos(x)-(m2g)(L/2)(cos(x))-fLsin(x)+N2Lcos(x)

I know that torque is the force multiplied by the distance to the pivot point. What I don't get is why we are only concerned with with each force's vertical distance from the pivot point and not the horizontal distance. Help, please?
 

Attachments

  • MTS_st_0.jpg
    MTS_st_0.jpg
    8.8 KB · Views: 1,112
Physics news on Phys.org
SA32 said:
I know that torque is the force multiplied by the distance to the pivot point.
Actually, torque is the force multiplied by the perpendicular distance to the pivot point--direction matters!
What I don't get is why we are only concerned with with each force's vertical distance from the pivot point and not the horizontal distance.
Since it's the perpendicular distance that counts: for vertical forces (like weight and normal force), it's the horizontal distance; but for horizontal forces (like friction), it's the vertical distance.
 
Ahh, that makes sense. Thanks!
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top