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Nataly [62]
4 years ago
10

Technician A says that tie rod ends are used to connect the steering linkage to the steering knuckles. Technician B says that so

me ball and socket joints are sealed and do not require lubrication. Which technician is correct
Physics
1 answer:
Iteru [2.4K]4 years ago
8 0

Answer:

Both technicians A and B are correct.                

Explanation:

The tie rod is present as a part  of a steering system. It consists of inner end as well as  outer end. It helps in transferring force from the steering link to the steering knuckle. And this causes to turn the wheel of the vehicle.

Usually a ball joint comes with pre-greased. The sealed ball and socket joint are lubed for their life, so they does not require lubrication.

Thus both the technicians are correct.

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Two vectors A and B are added to give a resultant R. The components of A are Ax = -8.0 units and Ay = 6.0 units and the componen
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Answer:

<em>10.09 units</em>

Explanation:

For the A

Ax = -8.0 units

Ay = 6.0 units

The resultant vector = Ra = \sqrt{A^2_{x} + A^2_{y}  }

Ra = \sqrt{(-8)^2 + 6^2  } =  10 units

For B

Bx = 1.0 units

By = -1.0 units

The resultant vector = Rb = \sqrt{B^2_{x} + B^2_{y}  }

Rb = \sqrt{1^2 + (-1)^2  } = \sqrt{2} units

Adding these two vectors A and B together, magnitude of vector R is

R = \sqrt{R^2_{a} + R^2_{b}  }

R =  \sqrt{10^2 + (\sqrt{2} ) ^2} = <em>10.09 units</em>

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A heavy mirror that has a width of 2 m is to be hung on a wall as shown in the figure below. The mirror weighs 700 N and the wir
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The translational equilibrium condition allows finding that the response for cable length with a maximum tension is

      L = 2.56 m

Newton's second law says that the force is proportional to the mass and the acceleration of the body, in the special case that the acceleration is zero, the relationship is called the translational equilibrium condition.

              ∑ F = 0

 

Where the bold indicates vectors, F is the force and the sum is for all external forces.

The reference systems are coordinate systems with respect to which the decomposition of the vectors is carried out and the measurements are made, in this case we will use a system with the horizontal x axis and the vertical y axis.

In the attachment we can see a free body diagram of the system, let's write the equilibrium condition for each axis.

x-axis  

         Tₓ -Tₓ = 0

y-axis

         T_y + T_y - W =0

         2T_y - W = 0

Let's use trigonometry to decompose the tension, we can see from the graph and the adjoint that each string is half the length, let's call the angle θ

          cos θ = \frac{T_x}{T}

          sin θ = \frac{T_y} { T}

          Tₓ = T cos θ

          T_y = T sin θ

We substitute

          2 T sin θ = W          (1)

The text indicates that the length of the block is 2 m, so the distance to the midpoint is

        x = 1 m

Let's use the Pythagoras' Theorem            

            H² = CA² + CO²

           CO = \sqrt{H^2 - CA^2}

           CA = x

           CO = \sqrt{(\frac{L}{2} )^2  - 1 }

Where CO is the opposite leg,  CA is the adjacent leg and H is the hypotenuse indicating H = L / 2,

Let's write the trigonometry functions

           sin θ = \frac{CO}{H}

Let's substitute        

            sin θ = \frac{2 \sqrt{\frac{L^2}{4} -1 } }{L}

Let's subtitute in the equation  1

          2 T  ( \frac{2 \sqrt{\frac{L^2}{4} -1 } }{L}   ) = W

          \sqrt{\frac{L^2}{4}-1  }  = \frac{1}{4}  \ \frac{W}{T} \ L

Let's solve by squaring

         \frac{L^2}{4}  -1 = \frac{1}{4} \ (\frac{W}{T} )^2 \ \frac{L^2}{4}  

         \frac{L^2}{4} ( 1 - \frac{1}{4} (\frac{W}{T})^2  ) -1 =0

They indicate that the maximum tension of the cable is T = 700N and the weight is worth W = 700N, we substitute the values

        \frac{L^2}{4} ( 1- \frac{1}{4}) - 1 =0 \\\frac{3 L^2}{16} = 1 \\ L^2 = \frac{16}{3} \\L = \sqrt{\frac{16}{3} }

        L =   2.31 m    

In conclusion using the translational equilibrium condition we can find that the response for cable length with maximum tension is

      L = 2.31 m

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