To determine the reactions at supports A and B, we will use the equations of static equilibrium.

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To determine the reactions at supports A and B, we will use the equations of static equilibrium.

Apply the equilibrium equation for the sum of vertical forces.
Assume upward forces are positive and downward forces are negative.
$\sum F_y = 0$ $R_A + R_B - 10 = 0$ $R_A + R_B = 10 \quad (*)$
Apply the equilibrium equation for the sum of moments about point A.
Assume counter-clockwise moments are positive.
$\sum M_A = 0$
The load of 10 units creates a clockwise moment about A.
The reaction $R_B$ creates a counter-clockwise moment about A.
$(R_B \times 6) - (10 \times 3) = 0$ $6 R_B - 30 = 0$ $6 R_B = 30$ $R_B = \frac{30}{6}$ $R_B = 5$
Substitute the value of $R_B$ into equation $(*)$ to find $R_A$ .
$R_A + 5 = 10$ $R_A = 10 - 5$ $R_A = 5$
The reactions at the supports are:
$R_A = 5$ units
$R_B = 5$ units

The final answer is $R_A = 5, R_B = 5$ .

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