Question 955490
CORRECTED ANSWER:
Here is a representation of the situation:
(I would assume that the box is uniformly dense, and would attach the rope to the middle of the side of the box that is facing up)
{{{drawing(500,500,-1,9,-4,6,
line(-1,0,9,0),triangle(0,0,8.19,0,8.19,5.74),
red(arc(0,0,2,2,-35,0)),locate(1,0.5,red(35^o)),
line(3.69,2.58,3.4,2.99),line(4.51,3.15,4.22,3.56),
line(3.4,2.99,4.22,3.56),line(3.69,2.58,4.51,3.15),
blue(triangle(3.95,3.07,3.95,0.17,5.31,1.13)),
blue(triangle(3.95,3.07,3.95,0.17,2.59,2.12)),
blue(line(3.95,3.07,5.31,1.13)),
blue(line(3.95,3.07,2.59,2.12)),
blue(arrow(3.95,3.07,5.31,1.13)),
blue(arrow(3.95,3.07,2.59,2.12)),
arrow(3.95,3.07,3.95,0.17),
red(arrow(3.95,3.07,3.13,2.5)),
red(line(3.95,3.07,3.13,2.5)),
green(arrow(3.95,3.07,6.13,4.60)),
green(line(3.95,3.07,6.13,4.60)),
red(arc(3.95,3.07,1.2,1.2,55,90)),
locate(3.98,1.5,W),locate(6.15,4.8,green(F)),
locate(2.3,2.8,blue(W[p])),locate(2.7,3.2,red(F[fr])),
locate(5.35,1.25,blue(N))
)}}} The force {{{green(F)}}} applied to make the box move uphill is represented by the green arrow.
The force of friction {{{red(F[fr])}}} opposing the movement is represented by the red arrow.
The weight of the box is represented by the black downwards arrow.
The weight of the box, {{{W}}} , can be decomposed into
{{{N}}} , a "normal" component, perpendicular to the surface of the incline
(here normal means perpendicular), and
{{{blue(W[p])}}} , a component parallel to the surface of the incline.
Those components of the weight are represented in blue.
The red arrow for the friction force is not drawn to scale for a good reason.
 
I assume that the forces are all expected to be measured in Newtons,
and that we can use {{{9.8}}}{{{"m"}}}{{{" /"}}}{{{s^2}}} for the acceleration of gravity.
That would make the magnitude of the weight (in Newtons)
{{{W=100*9.8=980}}} .
The magnitudes of all the other forces will be related to {{{W}}} by some factor.
The incline makes a {{{red(35^o)}}} angle with the a horizontal surface (real or imaginary),
so the normal weight component {{{blue(N)}}} makes a {{{red(35^o)}}} angle with the vertical weight {{{W}}} .
(It's thew same angle rotated by {{{90^o}}} and shifted).
The blue right triangles shown in the decomposition of {{{W}}} tell us that
{{{blue(N)=W*cos(35^o)=980*cos(35^o)=about980*0.819152}}} and
{{{blue(W[p])=W*sin(35^o)=980*sin(35^o)=about980*0.573576}}} .
The force of friction is {{{0.65}}} times {{{blue(N)}}} , so
{{{red(F[fr])=0.65*980*cos(35^o)=about0.65*980*0.819152}}} .
We do not have to worry about {{{blue(N)}}} any longer, because unless the box is sinking on a soft surface, the surface is applying an opposing force that "neutralizes" {{{blue(N)}}} .
As the box is moving at a constant speed, the sum of the other forces is zero,
so the magnitude of uphill-pointing {{{green(F)}}} is
the sum of the magnitudes of downhill-pointing {{{blue(W[p])}}} and {{{red(F[fr])}}} .
The approximate value of the magnitude of the force applied by the man is
{{{green(F)=980*0.573576+0.65*980*0.819152=980*(0.573576+0.65*0.81915)=980*(0.573576+0.532449)=980*1.106025=about}}}{{{highlight(1084)}}} .
I would not use any more significant digits for the result,
because I am using a generic {{{g=9.8}}}{{{"m"}}}{{{" /"}}}{{{s^2}}}  ,
{{{g}}} varies with latitude, from {{{9.78}}} at the Equator to {{{9.83}}} at the pole,
and I do not know where the man and the box are.
 
NOTE: If I had drawn to scale the red arrow for{{{red(F[fr])=about0.65*980*0.819152=about980*0.532449}}} ,
it would have been about the same size as the blue arrow for
{{{blue(W[p])=about980*0.573576}}} ,
and that would have made both superimposed arrows hard to see.