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- (a) The center of mass is given by
x com = [0 + 0 + 0 + ( m )(2.00) + ( m )(2.00) + ( m )(2.00)]/6.00 m = 1.00 m.
(b) Similarly, y com = [0 + ( m )(2.00) + ( m )(4.00) + ( m )(4.00) + ( m )(2.00) + 0]/6 m = 2.00 m.
(c) Using Eq. 12-14 and noting that the gravitational effects are different at the different locations in this problem, we have
x cog =
x 1 m 1 g 1 + x 2 m 2 g 2 + x 3 m 3 g 3 + x 4 m 4 g 4 + x 5 m 5 g 5 + x 6 m 6 g 6 m 1 g 1 + m 2 g 2 + m 3 g 3 + m 4 g 4 + m 5 g 5 + m 6 g 6 = 0.987 m.
(d) Similarly, y cog = [0 + (2.00)( m )(7.80) + (4.00)( m )(7.60) + (4.00)( m )(7.40) + (2.00)( m )(7.60) + 0]/(8.00 m + 7.80 m + 7.60 m + 7.40 m + 7.60 m + 7.80 m ) = 1.97 m.
- From
τ = r × F , we note that persons 1 through 4 exert torques pointing out of the
page (relative to the fulcrum), and persons 5 through 8 exert torques pointing into the page.
(a) Among persons 1 through 4, the largest magnitude of torque is (330 N)(3 m) = 990 N·m, due to the weight of person 2.
(b) Among persons 5 through 8, the largest magnitude of torque is (330 N)(3 m) = 990 N·m, due to the weight of person 7.
- The situation is somewhat similar to that depicted for problem 10 (see the figure that
accompanies that problem). By analyzing the forces at the “kink” where
F is exerted, we
find (since the acceleration is zero) 2 T sin θ = F , where θ is the angle (taken positive)
between each segment of the string and its “relaxed” position (when the two segments are
collinear). Setting T = F therefore yields θ = 30º. Since α = 180º – 2θ is the angle
between the two segments, then we find α = 120º.
- Three forces act on the sphere: the tension force
T of the rope (acting along the rope),
the force of the wall FN
(acting horizontally away from the wall), and the force of gravity mg
(acting downward). Since the sphere is in equilibrium they sum to zero. Let θ be the
angle between the rope and the vertical. Then, the vertical component of Newton’s
second law is T cos θ – mg = 0. The horizontal component is FN – T sin θ = 0.
(a) We solve the first equation for the tension: T = mg / cos θ. We substitute
cos θ = L / L^2^ + r^2 to obtain
(^2 2) (0.85 kg)(9.8 m/s )^2 (0.080 m)^2 (0.042 m) 2 9.4 N 0.080 m
mg L r T L
(b) We solve the second equation for the normal force: FN = T sin θ.
Using sin θ = r / L^2^ + r^2 , we obtain
2 2 2 2 2 2 2
(0.85 kg)(9.8 m/s )(0.042 m) 4.4 N. N (0.080 m)
Tr mg L r r mgr F L r L^ L r L
- We take the force of the left pedestal to be F 1 at x = 0, where the x axis is along the diving board. We take the force of the right pedestal to be F 2 and denote its position as x = d. W is the weight of the diver, located at x = L. The following two equations result from setting the sum of forces equal to zero (with upwards positive), and the sum of torques (about x 2 ) equal to zero:
1 2 1
F F W
F d W L d
(a) The second equation gives
1
3.0 m (580 N)= 1160 N 1.5 m
L d F W d
which should be rounded off to F 1 (^) = −1.2 × 10 3 N. Thus, | F 1 | = 1.2 × 10 3 N.
(b) Since F 1 is negative, indicating that this force is downward.
(c) The first equation gives F 2 (^) = W − F 1 =580 N+1160 N=1740 N
which should be rounded off to F 2 (^) = 1.7 × 103 N. Thus, | F 2 | = 1.7 × 103 N.
(d) The result is positive, indicating that this force is upward.
(e) The force of the diving board on the left pedestal is upward (opposite to the force of the pedestal on the diving board), so this pedestal is being stretched.
(f) The force of the diving board on the right pedestal is downward, so this pedestal is being compressed.
- Our notation is as follows: M = 1360 kg is the mass of the automobile; L = 3.05 m is the horizontal distance between the axles; " = (3.05 −1.78) m =1.27 mis the horizontal
distance from the rear axle to the center of mass; F 1 is the force exerted on each front wheel; and, F 2 is the force exerted on each back wheel.
(a) Taking torques about the rear axle, we find
2 3 1
(1360 kg) (9.80 m/s ) (1.27 m) 2.77 10 N. 2 2(3.05 m)
Mg F L
= = = ×
(b) Equilibrium of forces leads to 2 F 1 + 2 F 2 = Mg , from which we
obtain F 2 = 389. × 103 N.
- The angle of each half of the rope, measured from the dashed line, is
tan 1 0.30 m 1.. 9.0 m
θ −^
Analyzing forces at the “kink” (where
F is exerted) we find
550 N (^) 8.3 10 N. 3 2sin 2sin1.
F
T
= = = ×
- The forces on the ladder are shown in the diagram below. F 1 is the force of the window, horizontal because the window is frictionless. F 2 and F 3 are components of the force of the ground on the ladder. M is the mass of the window cleaner and m is the mass of the ladder.
The force of gravity on the man acts at a point 3.0 m up the ladder and the force of gravity on the ladder acts at the center of the ladder. Let θ be the angle between the
ladder and the ground. We use cos θ = d / L or sin θ = L^2^ − d^2 / L to find θ = 60º. Here
L is the length of the ladder (5.0 m) and d is the distance from the wall to the foot of the ladder (2.5 m).
(a) Since the ladder is in equilibrium the sum of the torques about its foot (or any other point) vanishes. Let " be the distance from the foot of the ladder to the position of the
window cleaner. Then, Mg " cos θ+ mg (^) ( L / 2 cos) θ− F L 1 sin θ= 0 , and
2 1 2
( / 2) cos [(75 kg) (3.0 m)+(10 kg) (2.5 m)](9.8 m/s ) cos 60 sin (5.0 m) sin 60 2.8 10 N.
M mL g F L
= ×
This force is outward, away from the wall. The force of the ladder on the window has the same magnitude but is in the opposite direction: it is approximately 280 N, inward.
(b) The sum of the horizontal forces and the sum of the vertical forces also vanish:
F F F Mg mg
1 3 2
The first of these equations gives F 3 (^) = F 1 = 2 8. × 102 N and the second gives
- The forces exerted horizontally by the obstruction and vertically (upward) by the floor are applied at the bottom front corner C of the crate, as it verges on tipping. The center of the crate, which is where we locate the gravity force of magnitude mg = 500 N, is a horizontal distance " = 0 375. m from C. The applied force of magnitude F = 350 N is
a vertical distance h from C. Taking torques about C , we obtain
(500 N) (0.375 m) 0.536 m. 350 N
mg h F
- (a) Analyzing the horizontal forces (which add to zero) we find Fh = F 3 = 5.0 N.
(b) Equilibrium of vertical forces leads to Fv = F 1 + F 2 = 30 N.
(c) Computing torques about point O , we obtain
( )( ) ( )( ) 2 3
10 N 3.0 m + 5.0 N 2.0 m = + = = 1.3m. 30 N
F d v F b F a d
- The (vertical) forces at points A, B and P are FA , FB and FP , respectively. We note that FP = W and is upward. Equilibrium of forces and torques (about point B ) lead to
F F W
bW aF
A B A
(a) From the second equation, we find
FA = bW / a = (15/5) W = 3 W = 3(900 N)=2.7 × 10 3 N.
(b) The direction is upward since FA > 0.
(c) Using this result in the first equation above, we obtain
4 4(900 N) 3.6 10 N^3
FB = W − FA = − W = − = − × ,
or | FB | = 3.6 × 10 N^3.
(d) FB points downward, as indicated by the minus sign.
- With pivot at the left end, Eq. 12-9 leads to
- m s g L 2 – Mgx + T (^) R L = 0
where m s is the scaffold’s mass (50 kg) and M is the total mass of the paint cans (75 kg). The variable x indicates the center of mass of the paint can collection (as measured from the left end), and T (^) R is the tension in the right cable (722 N). Thus we obtain x = 0.702 m.
- With pivot at the left end of the lower scaffold, Eq. 12-9 leads to
- m 2 g L 2 2 –^ mgd^ +^ T^ R^ L^2 = 0
where m 2 is the lower scaffold’s mass (30 kg) and L 2 is the lower scaffold’s length (2. m). The mass of the package ( m = 20 kg) is a distance d = 0.50 m from the pivot, and T (^) R is the tension in the rope connecting the right end of the lower scaffold to the larger scaffold above it. This equation yields T (^) R = 196 N. Then Eq. 12-8 determines T (^) L (the tension in the cable connecting the right end of the lower scaffold to the larger scaffold above it): T (^) L = 294 N. Next, we analyze the larger scaffold (of length L 1 = L 2 + 2 d and mass m 1 , given in the problem statement) placing our pivot at its left end and using Eq. 12-9:
- m 1 g L 1 2 –^ T^ L^ d^ –^ T^ R^ ( L^1 –^ d ) +^ T L^1 = 0.
This yields T = 457 N.
- We consider the wheel as it leaves the lower floor. The floor no longer exerts a force on the wheel, and the only forces acting are the force F applied horizontally at the axle, the force of gravity mg acting vertically at the center of the wheel, and the force of the step corner, shown as the two components fh and f (^) v. If the minimum force is applied the wheel does not accelerate, so both the total force and the total torque acting on it are zero.
We calculate the torque around the step corner. The second diagram indicates that the distance from the line of F to the corner is r – h , where r is the radius of the wheel and h is the height of the step.
The distance from the line of mg to the corner is r^2 r h rh h (^2 )
+ b − g = 2 −.
Thus F r b − h g − mg 2 rh − h^2 = 0. The solution for F is
2
2 2 2 2 2 2 2
2(6.00 10 m)(3.00 10 m) (3.00 10 m) (0.800 kg)(9.80 m/s ) (6.00 10 m) (3.00 10 m) 13.6 N.
rh h F mg r h − − − − −
× × − ×
× − ×
