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four masses a, b, c and d resolve at equal radii and are equally spaced along a shaft. the mass b is 7kg and the radii of c and d makes angles ot 90° and 240° respectively with the radius of b. find the magnitude of the masses a, c and d and the angular position of a so that the system maybe completely balanced​

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The net force in the radial direction must be zero to balance the system. This means that the sum of the forces in the x and y directions must be zero. We can write the equations as follows:

ΣFx = ma_r = 0

ΣFy = ma_θ = 0

where a_r and a_θ are the radial and tangential accelerations, respectively. The tangential acceleration is zero because the system is in equilibrium.

Let M be the total mass of the system. Then, the magnitude of mass a can be found using the equation:

Ma_r = Mb(a+b)sinθ

where θ is the angle between the radii of masses b and a. Since the system is balanced, we have:

Ma_r = Mb(a+b)sinθ = 0

Since Mb ≠ 0 and sinθ ≠ 0, we must have a = -b. This means that mass a must be 7 kg.

Next, we can find the magnitude of mass c using the equation:

Mc(a+c)sin(90°-θ) = Mb(b+c)sinθ

Substituting the values, we get:

Mc(a+c) = Mb(b+c)cosθ

Mc(a+c) = 7(b+c)cosθ

Similarly, we can find the magnitude of mass d using the equation:

Md(a+d)sin(θ-240°) = Mb(b+d)sinθ

Substituting the values, we get:

Md(a+d) = Mb(b+d)cos(θ-240°)

Md(a+d) = 7(b+d)cos(θ-240°)

Finally, to find the angular position of mass a, we can use the equation:

ΣFy = Ma_θ + Mb(b+a)cosθ + Mc(c+a)cos(90°-θ) + Md(d+a)cos(θ-240°) = 0

Substituting the values, we get:

7a + 14cosθ + 7c - 7dcosθ = 0

a + 2cosθ + c - dcosθ = 0

This equation can be solved numerically to find the value of θ.

User Christian Flem
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