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sukhopar [10]
3 years ago
13

They will only see the door move when both people exert the same force so that the forces are unbalanced. neither person exerts

a force on the door so that the net force is zero. one person exerts more force than the other so that the forces are unbalanced. both people exert the same force so that the forces are balanced.
Physics
1 answer:
Veseljchak [2.6K]3 years ago
5 0

Answer:

The correct answer is when one person exerts more force than the other so that the forces are unbalanced.

Explanation:

In fact, the door is initially at rest. In order to move the door, a net force different from zero should be applied, according to Newton's second law:

                                       <em> ∑F = ma</em>

where the term on the left is the resultant of the forces acting on the door, m is the door mass and a its acceleration. In order to move the door, the acceleration must be different from zero. But this means that the resultant of the forces acting on it must be different from zero: this is possible only if  the forces applied by the two persons are unbalanced, i.e. one person exerts more force than the other.

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Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. the bolts are slightly undersized and
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Answer:

The maximum shear stress in shaft AB, T_{ABmax} is 15 MPa

The maximum shear stress in shaft CD,  T_{CDmax} is 45.9 MPa

Explanation:

The formula for a shaft polar moment of inertia, J is given by  

J = \pi \times \frac{D^4}{32} =\pi \times \frac{r^4}{2}

Therefore, we have

J_{AB} = \pi \times \frac{D_{AB}^4}{32} =\pi \times \frac{r_{AB}^4}{2}

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D_{AB} = Diameter of shaft AB = 30 mm = 0.03 m

r_{AB} = Radius of shaft AB = 15 mm = 0.015 m

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and

J_{CD} = \pi \times \frac{D_{CD}^4}{32} =\pi \times \frac{r_{CD}^4}{2}

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D_{CD} = Diameter of shaft CD = 36 mm = 0.036 m

r_{CD} = Radius of shaft CD = 18 mm = 0.018 m

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J_{CD} = \pi \times \frac{0.036^4}{32} =\pi \times \frac{0.018^4}{2} = 1.65 × 10⁻⁷ m⁴

Given that the shaft AB and CD are rotated 1.58 ° relative to each other, we have;

1.58 °= 1.58 \times \frac{2\pi }{360} rad = 2.76 × 10⁻² rad.

That is \phi_r = 2.76 × 10⁻² rad.

However  \phi_r =  \phi_{C/D} -  \phi_{B/A}  

Where:

\phi_{B/A} = \frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G} and

\phi_{C/D} = \frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G}

T_{AB} and T_{CD}= Torque on shaft AB and CD respectively

T_{AB}  = Required

T_{CD}= 500 N·m

L_{AB} and L_{CD} = Length of shafts AB an CD respectively

L_{AB}  = 600 mm = 0.6 m

L_{CD} = 900 mm = 0.9 m

G = Shear modulus of the material = 77.2 GPa

Therefore;

\phi_r =  \phi_{C/D} -  \phi_{B/A}  =\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}

2.76 × 10⁻² rad =\frac{T_{CD}\cdot L_{CD}}{J_{CD} \cdot G} -\frac{T_{AB}\cdot L_{AB}}{J_{AB} \cdot G}

=\frac{500\cdot 0.9}{1.65 \times 10^{-7} \cdot 77.2\times 10^9} -\frac{T_{AB}\cdot 0.6}{7.95\times 10^{-7} \cdot 77.2\times 10^9}

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T_{AB} =  79.54 N.m

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Therefore T_{CD total } = 500 - 79.54 = 420.46 N·m

τ_{max} = \frac{T\times R}{J}

\tau_{ABmax} = \frac{T_{AB}\times R_{AB}}{J_{AB}} =  \frac{79.54\times 0.015}{7.95\times 10^{-8}} = 15 MPa

\tau_{CDmax} = \frac{T_{CD}\times R_{CD}}{J_{CD}} = \frac{420.46\times 0.018}{1.65\times 10^{-7}} = 45.9 MPa

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