The maximum acceleration the truck can have so that the refrigerator does not tip over is 4.15 m/s².
<h3>What will be the maximum acceleration of the truck to avoid tipping over?</h3>
The maximum acceleration is obtained by taking clockwise moments about the tipping point of rotation.
Clockwise moment = Anticlockwise moment
Ft * 1.58 m = F * 0.67 m
where
- Ft is tipping force = mass * acceleration, a
- F is weight = mass * acceleration due to gravity, g
m * a * 1.58 = m * 9.81 * 0.67
a = 4.15 m/s²
The maximum acceleration the truck can have so that the refrigerator does not tip over is 4.15 m/s².
In conclusion, the acceleration of the truck is found by taking moments about the tipping point.
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Will this one work?...................
Answer:
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The answer should be 5 it’s easy
Answer:
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Explanation:
For this problem let's use Newton's second law applied to each body
Body A
X axis
T = m_A a
Axis y
N- W_A = 0
Body B
Vertical axis
W_B - T = m_B a
In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension
We write the equations
T = m_A a
W_B –T = M_B a
We solve this system of equations
m_B g = (m_A + m_B) a
a = m_B / (m_A + m_B) g
In this initial case
m_A = M
m_B = M
a = M / (1 + 1) M g
a = ½ g
Let's find the tension
T = m_A a
T = M ½ g
T = ½ M g
Now we change the mass of the second block
m_B = 2M
a = 2M / (1 + 2) M g
a = 2/3 g
We seek tension for this case
T’= m_A a
T’= M 2/3 g
Let's look for the relationship between the tensions of the two cases
T’/ T = 2/3 M g / (½ M g)
T’/ T = 4/3
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e