Department of Natural Sciences
Clayton College & State University
November 21, 2005
Physics 1111 – Quiz 9
Name ___SOLUTION__________________________________
Pictured below is a very light wooden plank with two masses, 10.0 kg each, on top of it.
Find the reaction forces at points A and B.
NAx = 0
NAy = NA
NBx = 0
NBy = NB
w1x = 0
w1y = - m g
w2x = 0
w2y = - m g
Fx = 0

Fy = 0
NA + NB - m g – m g = 0
NA + NB = 2 m g
= 0
NA (3.00 m) - m g (2.25 m) = 0
NA = 73.5 N
NB = 122.5 N
Department of Natural Sciences
Clayton College & State University
November 20, 2006
Physics 1111 – Quiz 7
Name _____________________________________
A cat named Feynman (of 4.50 kg mass) is sitting on a horizontal bar supported by two
vertical cables. The length of the bar is 2.00 m and its mass is 10.0 kg. If Feynman sits at
a distance of 80.0 cm from the left cable,
a.
What is the tension in the left cable?
b. What is the tension in the right cable?
TLx = TL cos (90o) = 0
TLy = TL sin (90o) = TL
TRx = TR cos (90o) = 0
TRy = TR sin (90o) = TR
WFx = wF cos (-90o) = 0
WFy = wF sin (-90o) = - mF g = - (4.50 kg)(9.81 m/s2) = - 44.1 N
wsx = ws cos (-90o) = 0
wsy = ws sin (-90o) = - ms g = - (10.0 kg)(9.81 m/s2) = - 98.1 N
Fx = 0 => 0 = 0
Fy = 0 => TR + TL – 44.1 N – 98.1 N – 39.2 N = 0
TR + TL= 142 N

Assuming that the axis of rotation is at the point where the left cable is attached to the
scaffold:
TL = 0
TR = TR (2.00 m)
wF = - (44.1 N) (0.800 m) = - 35.3 N-m
sp = - (98.1 N) (1.00 m) = - 98.1 N-m
 = 0
TR (2.00 m) – 35.3 N-m – 98.1 N-m = 0
TR = 66.7 N
TL= 142 N – 66.7 N = 75.3 N
Clayton College & State University
Department of Natural Sciences
April 19, 2006
Physics 1111 – Quiz 10
Name _____SOLUTION______________________________
A uniform meter stick with a mass of 180 g is supported horizontally by two vertical
strings, one at the 0-cm mark and the other at the 90-cm mark. What is the tension in the
string (a) at 0 cm? (b) at 90 cm?
Translational equilibrium requirements:
TLx = TL cos (90o) = 0
TLy = TL sin (90o) = TL
TRx = TR cos (90o) = 0
TRy = TR sin (90o) = TR
wx = (mg) cos (-90o) = 0
wy = (mg) sin (-90o) = - mg = - (0.180 kg)(9.81 m/s2) = 1.77 N
Fx = 0: 0 = 0

Fy = 0:
TL + TR – 1.77 N = 0
TL + TR = 1.77 N
Rotational equilibrium requirements:

 TL = 0

 TR = + TR (0.900 m)

 w = - (1.77 N)(0.500 m) = -0.885 N-m

= 0
TR (0.900 m) – 0.885 N-m = 0
TR = 0.983 N
TL= 1.77 N – 0.983 N = 0.786 N
Department of Natural Sciences
Clayton State University
April 23, 2008
Physics 1111 – Quiz 12
Name ____SOLUTION_________________________________
Pictured below is a very light wooden plank with two masses, 10.0 kg each, on top of it.
Find the reaction forces at points A and B.
NAx = NA cos (90.0o) = 0
NAy = NA sin (90.0o) = NA
NBx = NB cos (90.0o) = 0
NBy = NB sin (90.0o) = NB
w1x = m g cos (-90.0o) = 0
w1y = m g sin (-90.0o) = - m g = - (10.0 kg)(9.81 m/s2) = -98.1 N
w2x = m g cos (-90.0o) = 0
w2y = m g sin (-90.0o) = - m g = - (10.0 kg)(9.81 m/s2) = -98.1 N

Fx = 0

Fy = 0
NA + NB – 98.1 N – 98.1 M = 0
NA + NB = 196 N
NA = 0
NB = NB (3.00 m)
w1 = 0
w2 = -w2 (2.00 m) = -(98.0 N) (2.00 m) = -196 N m
 = 0
NB (3.00 m) -196 N m = 0
NB = 65.4 N
NA + NB = 196 N
NB = 131 N
Department of Natural Sciences
Clayton State University
July 25, 2005
Physics 1111 – Quiz 10
Name ____SOLUTION_________________________________
Pictured below is a very light wooden plank with two masses, 10.0 kg each, on top of it.
Find the reaction forces at points A and B.
NAx = NA cos (90o) = 0
NAy = NA sin (90o) = NA
NBx = NB cos (90o) = 0
NBy = NB sin (90o) = NB
w1x = w1 cos (-90o) = 0
w1y = w1 sin (-90o) = - w1 = - m g = - (10.0 kg)(9.80 m/s2) = - 98.0 N
w2x = w2 cos (-90o) = 0
w2y = w2 sin (-90o) = - w2 = - m g = - (10.0 kg)(9.80 m/s2) = - 98.0 N
Fx = 0 => 0 = 0
Fy = 0 => NA + NB - 98.0 N - 98.0 N = 0
NA + NB = 196 N
NA = - NA (3.00 m)
NB = 0
w1 = w1 (2.25 m) = (98.0 N) (2.25 m) = 220.5 N m
w2 = 0
 = 0
- NA (3.00 m) + 220.5 N m = 0
NA = 73.5 N
NA + NB = 196 N
(73.5 N) + NB = 196 N
NB = 122.5 N
Department of Natural Sciences
Clayton State University
July 23, 2007
Physics 1111 – Quiz 11
Name __SOLUTION___________________________________
1. A person weighting 800 N stands with one foot on each of two bathroom scales.
Which statement is definitely true?
a. Each scale will read 800 N.
b. Each scale will read 400 N.
c. If one scale reads 500 N, the other will read 300 N.
d. None of the above is definitely true.
2. What condition or conditions are necessary for static equilibrium?
a. Fx = 0.
b. Fx = 0, Fy = 0,  = 0.
c.  = 0.
d. Fx = 0, Fy = 0.
3. The disk in the Figure below has two forces of equal magnitudes but opposite
directions acting on it. The disk is in
a. Translational equilibrium.
b. Rotational equilibrium.
c. Static equilibrium.
d. The disk is not in equilibrium.