Answer:
b. -11.6 cm
Explanation:
We have given parameters:
Length, l = 1.5 m = 150 cm
Mass of weight, 
= 20 kg
Width, x = 4 cm
Distance d = 4 cm
Mass of bar, 
= 5 kg
We are asked to find the center of mass from the mid-point, 
Since 3 weights are on the left and 2 weights are on the right, we know:
= 3 * 20 = 60 kg
= 2 * 20 = 40 kg
And also we know that, 
= 150/2 = 75 cm
For the left side, center of mass is:
cm
From the midpoint, the distance to the left is:
cm
For the right side, center of mass is:
cm
From the midpoint, the distance to the right will be:
cm
Hence,
cm
I assume the 100 N force is a pulling force directed up the incline.
The net forces on the block acting parallel and perpendicular to the incline are
∑ F[para] = 100 N - F[friction] = 0
∑ F[perp] = F[normal] - mg cos(30°) = 0
The friction in this case is the maximum static friction - the block is held at rest by static friction, and a minimum 100 N force is required to get the block to start sliding up the incline.
Then
F[friction] = 100 N
F[normal] = mg cos(30°) = (10 kg) (9.8 m/s²) cos(30°) ≈ 84.9 N
If µ is the coefficient of static friction, then
F[friction] = µ F[normal]
⇒ µ = (100 N) / (84.9 N) ≈ 1.2
<h2>RED!</h2><h3></h3><h3>On the visible spectrum, red has the lowest frequency.</h3><h3>(I'm an amateur astronomer, so I would know.)</h3>
