A student is using a 12.9 ft ramp to raise an object 6 ft above the ground.

Engineering

1 answer:

Hatshy [7]3 years ago

8 0

Answer:

2.150

Explanation:

The ideal mechanical advantage of the ramp is given as a ratio between the height of the ramp and the distance from the ground ie

Ideal mechanical advantage of the ramp= $\frac {12.9}{6}=2.150$

Therefore, the answer is 2.150 to the nearest thousandth

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determine the position d of the 6- kn load so that the average normal stress in each rod is the same.

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The load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2}$ area.

<h3>What is meant by torque?</h3>

The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is torque.

Let the beam is of length L

Now the stress on both the end is the same now we can say that torque on the beam due to two forces must be zero

$N_1 * x=N_2 *(L-x)$

also, we know that stress at both ends are same

$\frac{N_1}{12}=\frac{N_2}{8}$

$2 * N_1=3 * N_2$

Now from two equations we have

$$\frac{3}{2} N_2 * x=N_2 *(L-x)$

solving the above equation we have

$$x=\frac{2}{5} L$

so the load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2}$ area.

The complete question is:

47. the beam is supported by two rods ab and cd that have cross-sectional areas of $$$12mm^2$ and $$$8mm^2$ , respectively. determine the position d of the 6-kn load so that the average normal stress in each rod is the same.