# Triangular Fr

```MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Physics Department
8.01 Physics I
Fall Term 2006
L01 Quiz #2
NAME
Problem 1: Triangular Frame
F
θ
F
L
L
2.26
2.26
(a)
(b)
A triangular frame consists of three identical rods, each of mass M and length L. It rests
upright on a horizontal surface with its lower right corner against a stop about which the
frame could pivot.
a) (30%) A horizontal force of magnitude F is applied to the upper corner of the frame as
shown in (a) above. What is the largest value of F may have without causing the frame
to pivot upward about the stop? [Hint: Where is the center of mass of the frame?]
Choose to calculate the torques about the pivot point so that contact forces drop out.
The center of mass of the frame is at its geometrical center.
k
F2
r2
r1
~τ1 = ~r1 &times; F~1 = r1,⊥ |F~1 | k̂
r2,
~τ2
r1,
F1
~τ1 + ~τ2
= (L/2)(3M g) k̂ = (3/2)M gL k̂
= ~r2 &times; F~2 = −r2,⊥ |F~2 | k̂
√
= −( 3L/2)F k̂
√
= [(3/2)M gL − 3LF/2] k̂ = 0
√
⇒
F = 3M g
b) (10%) Imagine that instead the force was applied at an angle θ with respect to the
horizontal as in (b) above. What is the minimum value of θ that assures the frame will
not pivot for any magnitude F of the applied force?
The torque due to the applied force (~τ2 in the above calculation) will always be negative,
and thus potentially capable of tilting the frame upward, as long as the force points
outside of the lower right hand tip of the triangle, the pivot point. Therefore, the angle
θ must be greater than 60 degrees for the frame to be stable for all magnitudes of F .
1
Problem 2: Masses and Pulleys
s
y3
y1
y2
d
m3
m1
m2
Three different masses – m1 , m2 and m3 – are suspended from the ceiling by four pulleys
and a string of fixed length. The pulleys are massless, exhibit no friction, and have the same
radius R. Two of the pulleys are fixed to the ceiling and two are free to move up or down.
m1 and m2 are suspended below their pulleys by a fixed distance d. The positions yi of the
masses increase downward from the axes of the fixed pulleys – a distance s below the ceiling –
as shown in the figure. The object of this problem is to find the four equations that determine
the acceleration of each of the three masses, ÿi ≡ d2 yi /dt2 , and the tension T in the string
after the masses have been released from rest. You do not have to solve these equations.
a) (15%) Draw the force diagram appropriate to m1 and find the dynamical equation for
ÿ1 . Indicate on the figure above the part of the system to which your force diagram
applies.
T
T
Fy = ma
m1 g − 2T = m1 ÿ1
m1g
b) (15%) Draw the force diagram appropriate to m2 and find the dynamical equation for
ÿ2 .
T
T
Fy = ma
m2 g − 2T = m1 ÿ2
m2g
2
c) (15%) Draw the force diagram appropriate to m3 and find the dynamical equation for
ÿ3 .
T
Fy = ma
m3 g − T = m1 ÿ3
m3g
d) (15%) Find an equation which expresses the relationship between the three accelerations.
The length of the string is constant, thus dL/dt = d2 L/dt2 = 0.
L = s + 2(y1 − d) + 2(y2 − d) + y3 + 4πR
0 = 2ẏ1 + 2ẏ2 + ẏ3
0 = 2ÿ1 + 2ÿ2 + ÿ3
3
```