A ladder of length 2L and mass M is positioned on level ground leaning against a wall such that the angle between the ladder and the horizontal is α. The coefficient of static friction between the ladder and the wall and between the ladder and the ground is μstatic = 0.65. The centre of mass of the ladder is halfway along it.
(c) For the ladder to be in mechanical equilibrium:
(i) Write down equations for the total x- and y-components of the 5 forces acting on the
ladder.
(ii) Consider torques about the centre of the ladder. In which direction (into or out of the page) does the torque due to each of the 5 forces force act?
(iii) Write down an equation for the sum of the torques about the centre of mass of the ladder.
(iv) Use your equation for the torques to derive an expression for tan α in terms of the magnitudes of the forces acting.
(d)(i) If the ladder is just on the point of slipping at both the upper and lower ends, what
can you say about the pair of forces acting at each of the top and bottom of the ladder?
(ii) Hence use this information, with the information from (c)(i) and the expression you have derived in (c)(iv) to calculate the minimum angle that the ladder can form with the ground in order for it not to slip.
My work on this would be meaningless without a sketch (and assumed directons of forces). Take a look at this sample problem http://physicstasks.eu/654/a-leaning-ladder
To answer part (c) of the question:
(i) The total x- and y-components of the 5 forces acting on the ladder are:
X-components: F_friction_bottom + F_friction_top = 0 (Since the ladder is in mechanical equilibrium, the net force in the x-direction is zero.)
Y-components: F_ground + F_wall + F_friction_bottom + F_friction_top - mg = 0 (The sum of vertical forces is zero as the ladder is in mechanical equilibrium.)
(ii) The torque due to each of the 5 forces about the center of the ladder:
- F_ground: Torque is into the page (negative direction) as it tends to rotate the ladder in that direction.
- F_wall: No torque about the center of the ladder since it acts through the center of mass.
- F_friction_bottom: Torque is out of the page (positive direction) as it tends to rotate the ladder away from the wall.
- F_friction_top: Torque is into the page (negative direction) as it tends to rotate the ladder towards the wall.
- mg: No torque about the center of the ladder since it acts through the center of mass.
(iii) The equation for the sum of the torques about the center of mass of the ladder is:
Torque_net = Torque_F_ground + Torque_F_friction_bottom + Torque_F_friction_top
(iv) To derive an expression for tan α in terms of the magnitudes of the forces, we can use the torque equation:
0 = R x F_friction_top - L x F_ground - L x F_friction_bottom
where R is the distance of the center of mass from the bottom of the ladder.
Now, let's move on to part (d) of the question:
(i) If the ladder is just on the point of slipping at both the upper and lower ends, it implies that the friction forces at both ends have reached their maximum possible values. Therefore, the pair of forces acting at each of the top and bottom of the ladder would be: F_friction_top = μ_static * N_top and F_friction_bottom = μ_static * N_bottom, where N_top and N_bottom are the normal forces at the top and bottom of the ladder, respectively.
(ii) To calculate the minimum angle that the ladder can form with the ground in order not to slip, we need to consider the friction forces at the top and bottom and use the information from part (c)(i) and the expression derived in part (c)(iv). By substituting the values of μ_static, N_bottom, N_top, and the expression for tan α, you can find the angle using algebraic manipulations.
Note: You will need to know the equations for the normal forces N_bottom and N_top, as they depend on the geometry of the ladder and the angle α, to calculate the forces accurately.
(c) (i) The total x-components of the forces acting on the ladder are:
- The gravitational force acting downward, which can be decomposed into components along the x and y directions. The x-component is zero since it acts vertically and has no horizontal component.
- The normal force exerted by the ground, which is perpendicular to the ground and has no x-component.
- The static friction force exerted by the ground, parallel to the ground, and has no x-component.
- The static friction force exerted by the wall, perpendicular to the wall, and has no x-component.
So, the equation for the total x-components of the forces is:
N_wall - F_friction_ground - F_friction_wall = 0
where N_wall is the normal force exerted by the wall, F_friction_ground is the static friction force exerted by the ground, and F_friction_wall is the static friction force exerted by the wall.
The total y-components of the forces acting on the ladder are:
- The gravitational force acting downward, which can be decomposed into components along the x and y directions. The y-component is mg.
- The normal force exerted by the ground, which is perpendicular to the ground and has no y-component.
- The static friction force exerted by the ground, parallel to the ground, and has no y-component.
- The static friction force exerted by the wall, perpendicular to the wall, and has a y-component of N_wall.
So, the equation for the total y-components of the forces is:
N_ground - F_friction_ground - mg + N_wall = 0
where N_ground is the normal force exerted by the ground, F_friction_ground is the static friction force exerted by the ground, and N_wall is the normal force exerted by the wall.
(ii) The torque due to each of the 5 forces acts into the page.
(iii) The sum of the torques about the center of mass of the ladder is:
τ = R × (N_ground - F_friction_ground - mg) + L × (N_wall - F_friction_wall)
where R is the distance from the center of mass to the point of contact with the ground, and L is the length of the ladder.
(iv) Using the equation for torques, we can derive an expression for tan α in terms of the magnitudes of the forces:
τ = 0 (since the ladder is in mechanical equilibrium)
R × (N_ground - F_friction_ground - mg) + L × (N_wall - F_friction_wall) = 0
Taking the magnitude of the torques and rearranging:
|R × (N_ground - F_friction_ground - mg)| = |L × (N_wall - F_friction_wall)|
Using the fact that R = L/2 and simplifying:
|N_ground - F_friction_ground - mg| = 2|N_wall - F_friction_wall|
Now, we can divide both sides of the equation by |N_ground - F_friction_ground - mg| to get:
1 = 2|N_wall - F_friction_wall| / |N_ground - F_friction_ground - mg|
And finally, using the fact that the magnitude of the frictional force is given by |F_friction| = μN, where μ is the coefficient of static friction, we can substitute the frictional forces:
1 = 2|(N_wall - μ_ground - mg)| / |N_ground - F_friction_ground - mg|