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kondor19780726 [428]
3 years ago
11

How does the magnitude of the normal force exerted by the ramp in the figure compare to the weight of the static block? The norm

al force is:______ a. greater than the weight of the block. b. possibly greater than or less than the weight of the block, depending on whether or not the ramp surface is smooth. c. equal to the weight of the block. d. possibly greater than or equal to the weight of the block, depending on whether or not the ramp surface is smooth. less than the weight of the block.
Physics
1 answer:
DIA [1.3K]3 years ago
8 0

Answer:

less than the weight of the block.

Explanation:

From the free body diagram, we get.

The normal force is N = Mg cosθ

The tension in the string is T = Mg sinθ

Wight of the block when the block is static, W = Mg

Now since the magnitude of cosθ is in the range of : 0 < cosθ < 1,

therefore, the normal force is less than the weight of the static block.

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storchak [24]
The vector c has a magnitude of 24.6m and it is in the negative y direction. Therefore
\vec{c} = - 24.6 \hat{j}

The vector b is 41.4° up from the x-axis. Therefore
\vec{b} = b[cos(41.4^{o}) \hat{i} + sin(41.4^{o}) \hat{j} ] =b(0.75\hat{i} + 0.6613 \hat{j})

The vector a is 27.7° up from the x-axis. Therefore
\vec{a} = a[cos(22.7^{o})\hat{i} + sin(27.7^{o})\hat{j}] =  a(0.8854\hat{i} + 0.4648\hat{j})

Because \vec{a} +\vec{b} + \vec{c} = 0, the sum of the x and y components should be zero. Therefore,
For the x-component,
0.8854a + 0.75b = 0
 or
 a + 0.847b = 0                            (1)
For the y-component,
0.4648a + 0.6613b - 24.6 = 0
 or
 a + 1.4228b = 52.926                (2)

Subtract (1) from (2).
0.5758b = 52.926
b = 91.917
a = -0.847b = -77.854

Answer:
The magnitude of vector a is -77.85 m
The magnitude of vector b is 91.92 m
5 0
3 years ago
Apply: The earth's gravity is pulling on you. Are you pulling on the earth? Explain your
coldgirl [10]

Answer:

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Explanation:

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fomenos

Answer:

v = 186.90\,\frac{m}{s}

Explanation:

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v = \sqrt{2\cdot g\cdot h}

v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.031\,m)}

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The third one sliding friction
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