I need help with the following question:::::::::::::::::::::::::::::: ----------
ID: 1415208 • Letter: I
Question
I need help with the following question::::::::::::::::::::::::::::::
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A pendulum has a length L = 1.30 m. It hangs straight down in a jet plane about to take off as shown by the dotted line in the figure.
As the jet accelerates uniformly during take-off, the pendulum deflects horizontally by D = 0.370 m to a new equilibrum postion. Calculate the magnitude of the plane's acceleration.
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Four blocks are on a horizontal surface.
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The blocks are connected by thin strings with tensions Tx, Ty, Tz. The masses of the blocks are A=40.0 kg, B=14.0 kg, C=13.0 kg, D=40.0 kg. Two forces, F1=51.0N and F2=83.0N act on the masses as shown. Assume that the friction between the masses and the surface is negligible and calculate the tension Ty.
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A frictionless pulley with zero mass is attached to the ceiling, in a gravity field of 9.81 m/s2. Mass M2 = 0.25 kg is observed to be accelerating downward at 1.8 m/s2. Calculate the mass M1.
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The coefficent of static friction between the floor of a truck and a box resting on it is 0.30. The truck is traveling at 80.1 km/hr. What is the least distance in which the truck can stop and ensure that the box does not slide?
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Explanation / Answer
1. a = g D / (L² - D² )
=9.8 x 0.370/(1.32-0.372) = 2.9 m/sec^2.
2. Given:
A=40.0 kg, B=14.0 kg, C=13.0 kg, D=40.0 kg.
and Two forces, F1=51.0N and F2=83.0N
51-Ty = 40 a...1
Ty- Tx = 14a ....2
Tx-Tz = 13a....3
Tz-83 = 40a....4
From 4 Tz = 40a+ 83
From 3 Tx = 13a +40a+ 83 = 53a+83
From 1 Ty = 14a+53a+83 = 67a+83
From 2 Tx = 97-67a
=>97-67a = 53a+83
=>a = 0.116m/sec^2
Ty = 67 x 0.116+83 = 90.81
3.)From the free body diagrams of the two masses:
m2(g) - T = (m2)(a)
T = m2(g - a)
where
m2 = 0.250 kg (given)
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
T = tension on the string
a = acceleration = 1.8 m/sec^2
For the first mass,
T - (m1)g = (m1)a
or
T = m1(a + g)
where
m1 = unknown mass
and all the other terms have been previously defined.
Since the 2 T's are equal,
m2(g - a) = m1(a + g)
Solving for "m1"
m1 = m2*(g - a)/(a + g)
Substituting values,
m1 = (0.250)(9.8 - 1.8)/(9.8 + 1.8)
m1 = 0.172 kg.
4.)if the box mass is m
the weight is mg
and the maximum friction force is 0.30mg
as acceleration is force over mass
a = F/m
the maximum acceleration the truck can experience is
a = 0.30mg / m
a = 0.30g
a = 0.30(9.81)
a = 2.94 m/s²
as the truck will be slowing down, this acceleration will be negative
a = -2.94 m/s²
kinematic equation
vf² = vi² + 2ad
d = (vf² - vi²) / 2a
vi = 80.1 (1000/3600) = 22.25 m/s
d = (0² - 22.25²) / 2(-2.94)
d = 84.2 m