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Mkey [24]
2 years ago
8

Question: For an 80-N squeeze on the handle of the pliers, determine the force F applied to the round rod b... For an 80-N squee

ze on the handle of the pliers, determine the force F applied to the round rod by each jaw. In addition, calculate the force supported by the pin at A.

Physics
1 answer:
balandron [24]2 years ago
6 0
<span>Force applied to round rod is 217 Newtons. Force applied to pin A is 297 Newtons. The pliers are basically a pair of simple levers. And like all levers, the product of length and force will remain a constant. So for the first part, we have the fulcrum at "A" The force applied is 80 N over a length of 95 mm giving us 80 * 95 = 7600. So the force applied to round rod will be X * 35 = 7600 And solving for X, gives us X * 35 = 7600 X = 217.1428571 N The answer to part 2 is almost as easy. But you need to rethink how the lever is. Instead of the fulcrum being at pin "A", we'll consider it to be at the round rod. So the lever arm that's getting the 80-N of force is now 35+95 = 130 mm long and the short level arm is still 35 mm. So 130 * 80 = 35 * X Solving for X 130 * 80 = 35 * X 10400 = 35 * X 297.1428571 = X So the pin is resisting 297 Newtons of force.</span>
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7. A local sign company needs to install a new billboard. The signpost is 30 m tall, and the ladder truck is parked 24 m away fr
wolverine [178]
<h2>Solution :</h2>

Here ,

• Height of sign post = 30 m

• Distance between signpost and truck = 24 m

Let the

• Top of signpost = A

• Bottom of signpost = B

• The end of truck facing sign post be = C

Now as we can clearly imagine that the ladder will act as an hypotenuse to the Triangle ABC .

Where

• AB = Height of signpost = 30 m

• BC = distance between both = 24 m

• AC = Minimum length of ladder

→ AC² = AB² + BC² ( As we can see AB is perpendicular to BC )

→ AC² = (30)² + (24)²

→ AC² = 900 + 576

→ AC² = 1476

→ AC = 38.41875

or AC apx = 38.42

So minimum height of ladder = 38.42

6 0
2 years ago
Block A, mass 250 g , sits on top of block B, mass 2.0 kg . The coefficients of static and kinetic friction between blocks A and
masha68 [24]

Answer:

  F = 69.3 N

Explanation:

For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by

               fr = μ N

We define a reference system parallel to the floor

block B  ( lower)

Y axis  

            N - W₁-W₂ = 0

            N = W₂ + W₂

            N = (M + m) g

X axis

              F -fr = M a

for block A (upper)

X axis

              fr = m a                 (2)

so that the blocks do not slide, the acceleration in both must be the same.

Let's solve the system by adding the two equations

             F = (M + m) a          (3)

             a =\frac{F}{ M+m}

the friction force has the formula

            fr = μ N

             fr = μ (M + m) g

let's calculate

            fr = 0.34 (2.0 + 0.250) 9.8

            fr = 7.7 N

we substitute in equation 2

             fr = m a

             a = fr / m

             a = 7.7 / 0.250

             a = 30.8 m / s²

we substitute in equation 3

             F = (2.0 + 0.250) 30.8

             F = 69.3 N

5 0
2 years ago
As a person pushes a box across a floor, the energy from the person’s moving arm is transferred to the box, and the box and the
san4es73 [151]

Answer:

conserved

Explanation:

During this process the energy is conserved

3 0
2 years ago
Read 2 more answers
A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound
NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

f_n=\frac{nv_s}{4L}

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz

hence, the frequency of the third overtone is 102 Hz

8 0
2 years ago
The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a
navik [9.2K]

Answer:

230

Explanation:

\omega = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J

Energy used in one acceleration

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J

Number of accelerations would be given by

n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4

So the number of complete accelerations is 230

8 0
2 years ago
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