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3 years ago

Determine the net torque on the 2.0-m-long uniform beam

Physics

2 answers:

Snowcat [4.5K]3 years ago

8 0

a. The net torque on the 2.0-m-long uniform beam about point C is 14 Nm anticlockwise

b. The net torque on the 2.0-m-long uniform beam about point P is 13 Nm clockwise

<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

$\large {\boxed {F = ma }$

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

<em>This problem is about </em><em>moment of force</em><em>!</em>

<u>Given:</u>

F₁ = 56 N

θ₁ = 32°

F₂ = 65 N

θ₂ = 45°

F₃ = 52 N

θ₃ = 58°

<u>Unknown:</u>

τ = ?

<u>Solution:</u>

<em>About Point C:</em>

$\Sigma \tau_C = F_1d_1 \sin \theta_1 + F_2d_2 \sin \theta_2 + F_3d_3 \sin \theta_3$

$\Sigma \tau_C = 56(1) \sin 32^o + 65(0) \sin 45^o - 52(1) \sin 58^o$

$\Sigma \tau_C = 29.675 + 0 - 44.099$

$\Sigma \tau_C \approx -14 \texttt{ Nm}$ <em>( anticlockwise )</em>

$\texttt{ }$

<em>About Point P:</em>

$\Sigma \tau_P = F_1d_1 \sin \theta_1 + F_2d_2 \sin \theta_2 + F_3d_3 \sin \theta_3$

$\Sigma \tau_P = 56(2) \sin 32^o - 65(1) \sin 45^o + 52(0) \sin 58^o$

$\Sigma \tau_P = 59.351 - 45.962 - 0$

$\Sigma \tau_P \approx 13 \texttt{ Nm}$ <em>( clockwise )</em>

$\texttt{ }$

<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441
Newton's Law of Motion: brainly.com/question/10431582
Example of Newton's Law: brainly.com/question/498822

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

Romashka [77]3 years ago

4 0

Answer:

14.42 Nm

Explanation:

Assuming the figure attached and taking moment about point C

Torque is the sum of all the torques about C.

The distance between force pairs is half the beam length hence

Torque = (56sin32) *(1m) +(65sin45) *(0m) -(52sin58) *(1m)

Torque= 29.675 - 0 - 44.098 = 14.42 Nm anticlockwise

Read more about tension

brainly.com/question/15880959

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4 0

2 years ago

A freight train car has a mass of 2,000 kilograms and an acceleration of 1.8 m/s/s. What is the average force behind that train

abruzzese [7]

Answer:

F = 3600 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of force must be equal to the product of mass by acceleration.

ΣF = m*a

where:

F = force [N]

m = mass = 2000 [kg]

a = acceleration = 1.8 [m/s^2]

Now replacing:

F = 2000*1.8

F = 3600 [N]

8 0

4 years ago

When light of wavelength 240 nm falls on a potassium surface, electrons having a maximum kinetic energy of 2.93 eV are emitted.

olchik [2.2K]

A) We want to find the work function of the potassium. Apply this equation:

E = 1243/λ - Φ

E = energy of photoelectron, λ = incoming light wavelength, Φ = potassium work function

Given values:

E = 2.93eV, λ = 240nm

Plug in and solve for Φ:

2.93 = 1243/240 - Φ

Φ = 2.25eV

B) We want to find the threshold wavelength, i.e. find the wavelength such that the energy E of the photoelectrons is 0eV. Plug in E = 0eV and Φ = 2.25eV and solve for the threshold wavelength λ:

E = 1243/λ - Φ

0 = 1243/λ - Φ

0 = 1243/λ - 2.25

λ = 552nm

C) We want to find the frequency associated with the threshold wavelength. Apply this equation:

c = fλ

c = speed of light in a vacuum, f = frequency, λ = wavelength

Given values:

c = 3×10⁸m/s, λ = 5.52×10⁻⁷m

Plug in and solve for f:

3×10⁸ = f(5.52×10⁻⁷)

f = 5.43×10¹⁴Hz

7 0

3 years ago

Four long wires are each carrying 6.0 A. The wires are located

Firdavs [7]

Answer:

$B_T=2.0*10^-5[-\hat{i}+\hat{j}]T$

Explanation:

To find the magnitude of the magnetic field, you use the following formula for the calculation of the magnetic field generated by a current in a wire:

$B=\frac{\mu_oI}{2\pi r}$

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

I: current = 6.0 A

r: distance to the wire in which magnetic field is measured

In this case, you have four wires at corners of a square of length 9.0cm = 0.09m

You calculate the magnetic field in one corner. Then, you have to sum the contribution of all magnetic field generated by the other three wires, in the other corners. Furthermore, you have to take into account the direction of such magnetic fields. The direction of the magnetic field is given by the right-hand side rule.

If you assume that the magnetic field is measured in the up-right corner of the square, the wire to the left generates a magnetic field (in the corner in which you measure B) with direction upward (+ j), the wire down (down-right) generates a magnetic field with direction to the left (- i) and the third wire generates a magnetic field with a direction that is 45° over the horizontal in the left direction (you can notice that in the image attached below). The total magnetic field will be:

$B_T=B_1+B_2+B_3\\\\B_{T}=\frac{\mu_o I_1}{2\pi r_1}\hat{j}-\frac{\mu_o I_2}{2\pi r_2}\hat{i}+\frac{\mu_o I_3}{2\pi r_3}[-cos45\hat{i}+sin45\hat{j}]$

I1 = I2 = I3 = 6.0A

r1 = 0.09m

r2 = 0.09m

$r_3=\sqrt{(0.09)^2+(0.09)^2}m=0.127m$

Then you have:

$B_T=\frac{\mu_o I}{2\pi}[(-\frac{1}{r_2}-\frac{cos45}{r_3})\hat{i}+(\frac{1}{r_1}+\frac{sin45}{r_3})\hat{j}}]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[(-\frac{1}{0.09m}-\frac{cos45}{0.127m})\hat{i}+(\frac{1}{0.09m}+\frac{sin45}{0.127m})]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[-16.67\hat{i}+16.67\hat{j}]\\\\B_T=2.0*10^-5[-\hat{i}+\hat{j}]T$

5 0

4 years ago