This result is independent of the length of the ladder because
L is cancelled in the second equilibrium condition,
[link] . No matter how long or short the ladder is, as long as its weight is 400 N and the angle with the floor is
our results hold. But the ladder will slip if the net torque becomes negative in
[link] . This happens for some angles when the coefficient of static friction is not great enough to prevent the ladder from slipping.
Check Your Understanding For the situation described in
[link] , determine the values of the coefficient
of static friction for which the ladder starts slipping, given that
is the angle that the ladder makes with the floor.
A swinging door that weighs
is supported by hinges
A and
B so that the door can swing about a vertical axis passing through the hinges
[link] . The door has a width of
and the door slab has a uniform mass density. The hinges are placed symmetrically at the door’s edge in such a way that the door’s weight is evenly distributed between them. The hinges are separated by distance
Find the forces on the hinges when the door rests half-open.
A 400-N swinging vertical door is supported by two hinges attached at points
A and
B .
Strategy
The forces that the door exerts on its hinges can be found by simply reversing the directions of the forces that the hinges exert on the door. Hence, our task is to find the forces from the hinges on the door. Three forces act on the door slab: an unknown force
from hinge
an unknown force
from hinge
and the known weight
attached at the center of mass of the door slab. The CM is located at the geometrical center of the door because the slab has a uniform mass density. We adopt a rectangular frame of reference with the
y -axis along the direction of gravity and the
x -axis in the plane of the slab, as shown in panel (a) of
[link] , and resolve all forces into their rectangular components. In this way, we have four unknown component forces: two components of force
and
and two components of force
and
In the free-body diagram, we represent the two forces at the hinges by their vector components, whose assumed orientations are arbitrary. Because there are four unknowns
and
we must set up four independent equations. One equation is the equilibrium condition for forces in the
x -direction. The second equation is the equilibrium condition for forces in the
y -direction. The third equation is the equilibrium condition for torques in rotation about a hinge. Because the weight is evenly distributed between the hinges, we have the fourth equation,
To set up the equilibrium conditions, we draw a free-body diagram and choose the pivot point at the upper hinge, as shown in panel (b) of
[link] . Finally, we solve the equations for the unknown force components and find the forces.
(a) Geometry and (b) free-body diagram for the door.
Solution
From the free-body diagram for the door we have the first equilibrium condition for forces:
