<span>tension, compression, and shearing and can i get brainliest plz</span>
0.120L + 2.345L = 2.465L = 4 significant figures in the answer
The force vector that has a magnitude of 12.0 N. and is oriented 60° to the left of the (y) has the followings components:
To solve this exercise the formulas and procedures we will use are:
- v(x) = v * cosine (angle)
- v(y) = v * sine (angle).
Where:
- v= magnitude of the vector
- v(x) = component of the vector on the (x) axis
- v(y) = component of the vector on the (y) axis
- angle = angle
Information about the problem:
- angle = 60º
- v = 12.0 N
- v(x)= ?
- v(y)= ?
Applying the formula of the component of the vector in the (x) axis we have:
v(x) = v * cosine (angle).
v(x) = 12.0 N * cosine (60º)
v(x) =6 N
Applying the formula of the component of the vector in the (y) axis we have:
v(y) = v * sine (angle)
v(y) = 12.0 N * sine (60º)
v(y) = 10.39 N
<h3>What is a vector?</h3>
It can be said to be a straight line described by a point (a) and (b) that has direction and sense.
Learn more about vector at: brainly.com/question/2094736
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Answer:
392 N
Explanation:
Draw a free body diagram of the rod. There are four forces acting on the rod:
At the wall, you have horizontal and vertical reaction forces, Rx and Ry.
At the other end of the rod (point X), you have the weight of the sign pointing down, mg.
Also at point X, you have the tension in the wire, T, pulling at an angle θ from the -x axis.
Sum of the moments at the wall:
∑τ = Iα
(T sin θ) L − (mg) L = 0
T sin θ − mg = 0
T = mg / sin θ
Given m = 20 kg and θ = 30.0°:
T = (20 kg) (9.8 m/s²) / (sin 30.0°)
T = 392 N