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One of the most common application of torque and force diagrams is in the case of static objects. That is, we want to determine forces and positions where those forces act so as to guarantee that objects remain at rest with no turning or translational motion. The following are examples of "greatest hits" for statics problems.
Solution: The forces acting on the ladder come from gravity,
static friction (the ladder is not moving), and contact
with the floor and the wall. Since no friction is acting
at the wall, only the normal force,
Fw,
is acting there. At the floor, the static friction
magnitude is unknown because its value can be anywhere
between 0 and the maximum possible value of
musFN.
We note that we have chosen the angle theta such that
theta = arctan(3/4) = 37° and we have chosen the
rotation axis for calculation of torques to be perpendicular
to the plane of the page and passing through the point of
contact between the ladder and the floor. Since we want
the translational and angular accelerations to be zero,
we need the net torque and net force to be zero, so
and
| x: | fs - Fw | = 0 ==> |
| fs | = Fw | |
| y: | FN - W | = 0 ==> |
| FN | = W = 60 Nt | |
| T: | Fw, pL - Wp(½ L) | = 0 ==> |
| Fwsin(theta) | = ½ Wcos(theta) ==> | |
| Fw | = ½ Wcot(theta) | |
| = ½ (60 Nt)cot(37° ) | ||
| Fw | = 40 Nt |
From our x equation of motion, we have
fs =
Fw = 40 Nt.
Solution: All of our force diagrams will appear the same. The
only things that have changed is that now x and y no
longer equal 3 m and 4 m, respectively. However, it
must still be true that (5 m)2 =
x2 + y2. Also, we see that the
frictional force does point to the right as we assumed
in the force diagram for the previous problem, so the
static frictional force will have to go to its maximum
value, mus
FN = (0.3)(60 Nt) = 18 Nt, to keep the
ladder from slipping. From our x equation of motion, we
have
To get the maximum distance of the bottom of the ladder
from the wall, we use the torque equation from the
previous problem to determine the maximum value of theta.
| Fw, max | = ½ Wcot(thetamax) ==> |
| thetamax | = arccot[2Fw, max/ W] |
| = arccot[2(18 Nt)/(60 Nt)] = 59° |
This maximum value of theta corresponds to
a maximum distance from the wall as follows:
Solution: We draw the free-body diagram depicting all the forces
and torques acting.
The torques should be calculated around an axis perpendicular
to the page and going through the connection between the
hinge and the beam. This eliminates the need to find the
torque due to the two unknown forces which the hinge provides
on the beam. For static conditions, the net forces and
torques must be zero, so
| x: | FNx - T cos(theta) | = 0 ==> |
| FNx | = T cos(theta) | |
| y: | T sin(theta) - FNy - W1 - W2 | = 0 ==> |
| FNy | = T sin(theta) - W1 - W2 | |
| torque: | (T sin(theta))L - W1(L/2) - W2L | = 0 ==> |
| T sin(theta) | = ½ W1 + W2 |
With T*sin(theta) known in terms of
W1 and
W2, we can
find T as (½ W1 +
W2)/sin(theta).
We also have FNx =
T cos(theta) = (½ W1 +
W2)cot(theta)
and
The minus sign indicates that we chose the wrong direction for FNy. This force actually points up rather than down. Keep in mind that the equations for force and torque will automatically indicate the correct directions for forces provide you are consistent in keeping track of signs.
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Up: Completing the Circle
Previous: Rolling Motion cont'd.
larryg@upenn5.hep.upenn.edu