A barge floating in fresh water (rho = 1000 kg/m3) is shaped like a hollow rectangular prism with base area A = 550 m2 and heigh
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- Let, the maximum height covered by projectile be
- Projectile is thrown with a velocity = v
- Angle of projection = θ
- Velocity of projectile at a height half of the maximum height covered be
______________________________
Then –
- Now, the vertical component of velocity of projectile at the height half of
will be –
Therefore, the vertical component of velocity of projectile at this height will be–
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Answer:
Fe= 28.2 N : Magnitude of the equilibrant (Fe)
β = 18.34° , clockwise from the positive x axis
Explanation:
Concept of the equilibrant
It is called equilibrant to a force with the same magnitude and direction as the resulting one (in case it is non-zero) but in the opposite direction. Adding vectorially to all the forces (that is to say the resulting one) with the equilibrant you get zero
To solve this problem we decompose the forces given into x-y components to find the resulting force:
Look at the attached graphic
F₁= 33.4 N , θ₁=23.8° clockwise from the positive y axis (y+)
F₁x= 33.4 *sin23.8° = 13.48 N
F₁y= 33.4 *cos23.8° =30.6 N
F₂=46.1 N , θ₂=28.8 counterclockwise from the negative x axis (x-)
F₂x= -46.1 *cos28.8° = -40.4 N
F₂y= -46.1 *sin28.8° = -22.2 N
Components of the resultant in x-y R(x,y)
Rx= 13.48 N -40.4 N = - 26.92 N
Ry= 30.6 N -22.2 N = + 8.4 N
Components of the equilibrant in x-y Fe(x,y)
Fex= +26.92 N
Fey= - 8.4 N
Magnitude of the equilibrant (Fe)
Fe= 28.2 N
Angle the equilibrant makes with the x axis ( β)
β = -18.34°
β = 18.34° , clockwise from the positive x axis
