patel (rpp467) – Quest #7 Statics (part1/3) – ha – (51122013) This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Consider an extended object (not a point), ~ acting on it, producing torques with forces F ~τ . Is it possible for a situation to exist in which the net force acting on the object (the net force is the sum of all the individual forces actX ~ ing on the object) is equal to zero F =0 while the net torque about any axis (the net torque is the sum of all the torques Xactingon the object) is not equal to zero ~τ 6= 0 ? 1. No. 2. Yes. correct Explanation: Yes — if the forces are equal and opposite, but do not act along a common line, like pulling the top of a box while pushing the bottom (on the same side), the net force is zero, while the net torque is not! 002 (part 2 of 2) 10.0 points Is it possible for a situation to exist in which the acting on the object is zero Xnet torque ~τ = 0 while the net force acting on the X ~ 6= 0 ? object is not equal to zero F
ladder is stable when µ ≥
L
h 45
N
◦
mg f
b
µ = 0.4
b
The ladder will be 1. stable. 2. unstable. correct 3. at the critical point of slipping. Explanation: Stability requires 1 = 0.5 . µ≥ 2 tan 45◦ Since µ = 0.04 < 0.5 , the ladder is unstable. 004 (part 1 of 3) 10.0 points A solid sphere of radius R and mass M is held against a wall by a string being pulled at an angle θ . f is the magnitude of the frictional force and W = M g . F θ R
2. No.
003 10.0 points A ladder is leaning against a smooth wall. There is friction between the ladder and the floor, which may hold the ladder in place; the
1 . 2 tan θ
F
1. Yes. correct
Explanation: Yes again — for example, just one force acting on the center of mass of an object produces no torque, but certainly produces a net force!
1
P
W To what does the torque equation
X
~τi = 0
i
about point O (the center of the sphere) lead?
patel (rpp467) – Quest #7 Statics (part1/3) – ha – (51122013)
2
cally, 1. F + W = f X
2. F = f correct
Fyi = F sin θ + f − W = 0
i
3. F cos2 θ = f
F sin θ + f = W .
4. F sin θ = f 5. F sin θ cos θ = f
006 (part 3 of 3) 10.0 points Find the smallest coefficient of friction µ needed for the wall to keep the sphere from slipping.
6. W = f Explanation:
1. µ = sin θ
F R
θ
R
2. µ = f P
1 tan θ
3. µ = tan θ 1 correct cos θ 1 5. µ = sin θ 4. µ =
W Applying rotational equilibrium about O, X the center of the sphere, ~τi = 0, so i
τCW = τCCW FR=fR F =f.
005 (part 2 of 3) 10.0 points To what does the vertical component of the force equation lead? 1. F sin θ + f = W correct
6. µ = cos θ Explanation: Let N be the normal force. f ≤ µ N ; when µ is minimal, f = µ N . Applying translational equilibrium horizontally,
X
Fxi = F cos θ − N = 0
i
µ N cos θ − N = 0 N (µ cos θ − 1) = 0 µ=
1 . cos θ
2. F sin θ = W 3. F sin θ = f 4. F sin θ = f + W 5. F cos θ + W = f Explanation: Applying translational equilibrium verti-
007 10.0 points The system shown in the figure is in equilibrium. A 14 kg mass is on the table. A string attached to the knot and the ceiling makes an angle of 50◦ with the horizontal. The coefficient of the static friction between the 14 kg mass and the surface on which it rests is 0.32.
patel (rpp467) – Quest #7 Statics (part1/3) – ha – (51122013)
◦
50
3
008 (part 1 of 2) 10.0 points The uniform diving board has a mass of 26 kg .
14 kg 1m m
A
3.3 m B
What is the largest mass m can have and still preserve the equilibrium? The acceleration of gravity is 9.8 m/s2 .
Find the force on the support A when a 72 kg diver stands at the end of the diving board. The acceleration of gravity is 9.81 m/s2 .
Correct answer: 5.33906 kg.
Correct answer: 2.62418 kN.
Explanation: Let : M = 14 kg , m = 5.33906 kg , and θ = 50◦ . For the system to remain in equilibrium, the net forces on both M and m should be zero, so the tension in the rope has an upper bound value Tmax , where Tmax cos θ = µ M g (1) µM g Tmax = cos θ (0.32) (14 kg) (9.8 m/s2 ) = cos 50◦ = 68.3025 N . For m to remain in equilibrium Tmax sin θ = mmax g (2) Tmax sin θ (68.3025 N) sin 50◦ mmax = = g 9.8 m/s2
Explanation:
Let :
Consider rotational equilibrium about B: FB xd A
= 5.33906 kg . Alternate Solution: Equations 1 and 2 come directly from the free-body diagram for the knot. Dividing Eq. 2 by Eq. 1, mmax tan θ = µM mmax = µ M tan θ = (0.32) (14 kg) tan 50◦ = 5.33906 kg .
m = 26 kg , M = 72 kg , ℓ = 4.3 m , x1 = 1 m , and xd = 3.3 m .
x1
x2
x3
B FA
x2 =
mg
Mg
ℓ 4.3 m − x1 = − 1 m = 1.15 m 2 2
patel (rpp467) – Quest #7 Statics (part1/3) – ha – (51122013) Applying rotational equilibrium at point B, X ~τ = FA x1 − m g x2 − M g xd = 0 x2 m + xd M FA = g x1 (1.15 m) (26 kg) + (3.3 m) (72 kg) = 1m 1 kN ×(9.81 m/s2 ) × 1000 N = 2.62418 kN .
Explanation: Let :
x≡
FB = FA + (m + M ) g = 2.62418 kN + (26 kg + 72 kg) (9.81 m/s2 ) 1 kN × 1000 N = 3.58556 kN .
and
n X
xi M
i=1 n X
M
n
1X = xi , n i=1
i=1
where xi is the CM position of the ith brick. The CM of a single brick is in its middle, so the maximum overhang is
Correct answer: 3.58556 kN. Explanation: Applying translational equilibrium, X Fy = FB − FA − m g − M g = 0
L = 27 m n = 1.
The center of mass (CM) of a stack of n bricks (of equal mass M ) is
The direction of FA is downward, so it is tension. 009 (part 2 of 2) 10.0 points Find the force on the support B at that same instant.
4
x=
1 1 L = (27 m) = 13.5 m . 2 2 27 m x
011 (part 2 of 4) 10.0 points Two identical uniform bricks of length 27 m are stacked over the edge of a horizontal surface. 27 m
The direction of FB is upward, so it is compression. 010 (part 1 of 4) 10.0 points A uniform brick of length 27 m is placed over the edge of a horizontal surface. 27 m x What maximum overhang is possible for the single brick (without tipping)? Correct answer: 13.5 m.
x What maximum overhang is possible for the two bricks (without tipping)? Correct answer: 20.25 m. Explanation: Let :
n = 2.
The bricks will just balance when the CM of the stack is placed at the fulcrum (the edge
patel (rpp467) – Quest #7 Statics (part1/3) – ha – (51122013) of the horizontal surface). Place the left edge of the new brick below the CM of the previous brick and measure from the left edge of the 1 top brick. The top brick can extend L 2 from the left edge of the brick under it, so the position of the CM of the new brick is x2 = x
so for the stack of 3 bricks, 1 1 5 11 x = 1+ + L= L 3 2 4 12 n=3 11 (27 m) = 24.75 m . = 12 27 m
1 1 1 + L = L+ L = L, n=1 2 2 2
and for the stack of 2 bricks, 1 3 1 1+ L= L x = 2 2 4 n=2 3 = (27 m) = 20.25 m . 4 27 m
5
x
013 (part 4 of 4) 10.0 points n identical uniform bricks of length 27 m are stacked over the edge of a horizontal surface. etc.
x
012 (part 3 of 4) 10.0 points Three identical uniform bricks of length 27 m are stacked over the edge of a horizontal surface. 27 m
x What maximum overhang is possible for the n bricks (without tipping)? n
LX 1 1. x = 2 2i− 1
x
i=1
What maximum overhang is possible for the three bricks (without tipping)?
n X i 2. x = L 2i
Correct answer: 24.75 m.
LX1 3. x = correct 2 i
i=1 n
i=1
Explanation:
4. x = L Let :
n = 3.
For three bricks, the position of the CM of the new brick is 3 1 5 1 + L = L+ L = L, x3 = x 2 4 2 4 n=2
n X i=1
5. x = L 6. x =
n X
1 i! + 1 1 i+1
i=1 n X
3L 2
i=1
1 i+2
patel (rpp467) – Quest #7 Statics (part1/3) – ha – (51122013) n X 1 7. x = L 2i i=1
n LX1 8. x = 2 i! i=1 n 1 1 LX + 9. x = 2 i+1 4
6
The pattern continues in this manner, so for n bricks, 1 1 1 1 1 x= 1+ + + +···+ L 2 2 3 4 n n 1 X1 . = L 2 i i=1
i=1
Explanation: Rewrite the first three solutions: 1 1 x = = (1) L 2 2 n=1 1 1 1 3 3 L= 1+ L = L= x 4 2 2 2 2 n=2 11 1 11 x = L L= 12 2 6 n=3 1 6 3 2 L + + = 2 6 6 6 1 1 1 = 1+ + L. 2 2 3 For four bricks, the CM of the new brick is 11 1 17 1 L+ L= L , so x4 = x + L= 2 12 2 12 n=3 1 1 5 17 = x 1+ + + L n=4 4 2 4 12 1 25 25 L L= = 24 2 12 1 12 6 4 3 = L + + + 2 12 12 12 12 1 1 1 1 1+ + + L. = 2 2 3 4 For five the CM of the new brick is bricks, 1 25 1 37 x5 = x + L= L+ L= L , so 2 24 2 24 n=4 1 5 17 37 1 1+ + + L + x = 5 2 4 12 24 n=5 137 1 137 = L L= 120 2 60 1 60 30 20 15 12 = L + + + + 2 60 60 60 60 60 1 1 1 1 1 1+ + + + L. = 2 2 3 4 5
etc.
x The maximum overhang only occurs when the top block is extended as far out as possible on the second block, and the top two blocks are extended as far out as possible on the third block, and so forth. Other mechanical arrangements do not give a maximum overhang.
