#### Answer

(a) The force at the support point is 1920 N.
(b) The force at the left-hand end is 1140 N.

#### Work Step by Step

(a) Let's consider an axis of rotation at the left end of the board. Note that the force at the left-hand end exerts zero torque about this axis. Let $F_s$ be the force of the support.
$\sum \tau = 0$
$(F_s)(1.00~m) - (280~N)(1.50~m)-(500~N)(3.00~m) = 0$
$(F_s)(1.00~m) = (280~N)(1.50~m)+(500~N)(3.00~m)$
$F_s = \frac{(280~N)(1.50~m)+(500~N)(3.00~m)}{1.00~m}$
$F_s = 1920~N$
The force at the support point is 1920 N.
(b) The sum of the vertical forces must equal zero. Let $F_L$ be the force at the left-hand end of the board.
$\sum F = 0$
$F_L+280~N+500~N-1920~N=0$
$F_L = 1920~N-280~N-500~N$
$F_L = 1140~N$
The force at the left-hand end is 1140 N.