Let consider the whole system to be at equilibrium where the net force is zero. The forces [tex]F_A[/tex] and [tex]F_B[/tex] are acting in the upward direction while the weight (w) of the crate is acting in the downward direction.
The force at equilibrium in the y-direction is computed as:
[tex]\sum F_y = ma_y[/tex]
[tex]F_A +F_B= mg[/tex]
[tex]\mathbf{F_A = mg -F_B}[/tex] ---- (1)
Recall that the weight w = mg
where;
- mass (m) = 200 g
- g(constant gravity) = 9.8 m/s²
w = 200 × 9.8
w = 1960 N
The torque about the bottom end is also zero. i.e.
[tex]\sum \tau=0[/tex]
∴
[tex]F_B L_Z - wL_w = 0[/tex] --- (2)
[tex]F_B L_Z= wL_w[/tex]
[tex]F_B = \omega (\dfrac{L_w}{L_z})[/tex]
[tex]F_B =1960\times (\dfrac{0.375\times cos 45^0}{1.25\times cos 45^0})[/tex]
[tex]F_B =1960\times (0.3)[/tex]
[tex]\mathsf{F_B = 588 \ N}[/tex]
Replacing the value for the force of B into equation (1); we have:
[tex]F_A = mg - F_B[/tex]
[tex]F_A = 1960 - 588[/tex]
[tex]\mathsf{F_A = 1372 N}[/tex]
Thus, we can conclude that the magnitude of these forces i.e force B and force A are 588 N and 1372 N respectively.
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