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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

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