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2 years ago

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While a roofer is working on a roof that slants at 38.0 ∘ above the horizontal, he accidentally nudges his 95.0 n toolbox, causi

ng it to start sliding downward, starting from rest. part a if it starts 4.60 m from the lower edge of the roof, how fast will the toolbox be moving just as it reaches the edge of the roof if the kinetic friction force on it is 20.0 n ?

1 answer:

Stolb23 [73]2 years ago

7 0

1. Find all the forces on the toolbox
2. Use the total force to find the total acceleration
3. Use the acceleration to find the final velocity by combining these equations:
$x = \frac{1}{2} at^2 =\ \textgreater \ t^2 = \frac{2x}{a} \\ v = at =\ \textgreater \ t^2 = \frac{v^2}{a^2} \\ \\ v = \sqrt{2xa}$

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san4es73 [151]

<span>When the particles of a medium move with simple harmonic motion, this means the wave is a sinusoidal wave.

Know that a sinusoidal curve can describe either sine or cosine functions (remember your cofunction identities for sine and cosine).</span>

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2 years ago

Read 2 more answers

A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w

Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, $a=3\ m/s^2$

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

$m=\dfrac{W}{g}$

$m=\dfrac{150}{9.8}$

$m=15.3\ kg$

Frictional force is given by :

$F=ma$

$F=15.3\times 3$

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

8 0

2 years ago

The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine th

Luba_88 [7]

Here is the complete question

The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine the horizontal shear stress at point H, which is located L = 20 mm below the centriod

The missing image which is the remaining part of this question is attached in the image below.

Answer:

The horizontal shear stress at point H is $\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

Explanation:

Given that :

The internal shear force V = 80 kN = 80 × 10³ N

The moment of inertia = 64,900,000

The length = 20 mm below the centriod

The horizontal shear stress $\tau$ can be calculated by using the equation:

$\tau = \dfrac{VQ}{Ib}$

where;

Q = moment of area above or below the point H

b = thickness of the beam = 10 mm

From the centroid ;

Q = $Q_1 + Q_{2}$

Q = $A_1y_1 + A_{2}y_{2}$

Q = ( ( 70 × 10) × (55) + ( 210 × 15) (90 + 15/2) ) mm³

Q = ( ( 700) × (55) + ( 3150 ) ( 97.5) ) mm³

Q = ( 38500 + 307125 ) mm³

Q = 345625 mm³

$\tau_H = \dfrac{VQ}{Ib}$

$\tau_H = \dfrac{80*10^3 * 345625}{64900000*10 }$

$\tau_H = \dfrac{2.765*10^{10}}{649000000 }$

$\tau_H = 42.60400616 \ N/mm^2$

$\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

The horizontal shear stress at point H is $\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

7 0

2 years ago

A 900 kg steel beam is supported by the two ropes shown in (Figure 1) . Calculate the tension in the rope.

Rzqust [24]

Let T1 and T2 be tension in ropes1 and 2 respectively.
<span>since system is stationary (equilibrium), considering both ropes + beam as a system </span>

<span>for horizontal equilibrium (no movement in that direction, so resultant force must be zero horizontally) </span>
<span>T1sin(20) = T2sin(30) </span>
<span>=> T1 = T2sin(30) / sin(20) </span>

<span>for vertical equilibrium, (no movement in this direction, so resultant force must be zero vertically) </span>
<span>T1cos(20) + T2cos(30) = mg </span>

<span>m = 900kg, substituting for T1 </span>
<span>T2sin(30)*cos(20)/sin(20) + T2cos(30) = 900g </span>
<span>2.328*T2 = 900*9.8 </span>
<span>T2 = 3788.65N </span>
<span>so T1 from (1) </span>
<span>T1 = 5535.21N</span>

8 0

2 years ago

Two window washers, Bob and Joe, are on a 3.00 m long, 395 N scaffold supported by two cables attached to its ends. Bob weighs 8

WINSTONCH [101]

Answer:

- the forces on the left hand side is 1.038 kN

- the forces on the right hand side is 1.483 kN

Explanation:

Given the data in the question, as illustrated in the image below;

Length of the scaffold = 3 m

weight of the scaffold = 395 N

Weight of Bob = 805 N and stands 1 m from the left end

weight of washing equipment = 500N and on sits 2 m from the left end

Weight of Joe = 820 N and stand 0.500 m from the right end

so the force on the left cable will be;

$T_{left$ = $\frac{1}{3m}$ [ (805 N)( (3-1) m) + ( 395 N )( $\frac{3}{2} m$ ) + ( 500 N )(1m ) + ( 820 N)( 0.500m ) ]

$T_{left$ = $\frac{1}{3m}$ [ 1610 + 592.5 + 500 + 410 ]

$T_{left$ = $\frac{1}{3m}$ [ 3112.5 ]

$T_{left$ = 1037.5 N

$T_{left$ = 1.038 kN

Therefore, the forces on the left hand side is 1.038 kN

On the right hand side;

$T_{Right$ = ( 805 N + 395 N + 500 N + 820 N ) - 1037.5 N

$T_{Right$ = 2520 N - 1037.5 N

$T_{Right$ = 1482.5 N

$T_{Right$ = 1.483 kN

Therefore, the forces on the right hand side is 1.483 kN

5 0

1 year ago