In the horizontal direction, the forces acting on the person are
• friction with magnitude <em>f</em>, opposing motion, and
• the horizontal component of the pulling force (itself with mag. <em>p</em> ) with mag. <em>p</em> cos(30º), in the direction of motion.
There is no friction in the vertical direction, so we omit any discussion of the vertical forces.
By Newton's second law, we then have
<em>p</em> cos(30º) - <em>f</em> = <em>m</em> <em>a</em> cos(30º)
where <em>m</em> is the person's mass, and <em>a</em> is their acceleration so that <em>a</em> cos(30º) is the magnitude of the horizontal component of acceleration. The person is pulled by a force of <em>p</em> = 401 N, so solve for <em>f</em> :
(401 N) cos(30º) - <em>f</em> = (53 kg) (0.59 m/s²) cos(30º)
<em>f</em> ≈ 320 N