Question

A symmetrical table of height 0.780 m, length 1.30 m, and weight 431 N is dragged across the floor by a force applied to its front edge. The force is directed to the right and upward and makes an angle of 30.0° with the horizontal.

Answer 1

here,
weight of table, w = 431 N
legth of table, l = 1.30 m
height of table, h = 0.780 m
angle, A = 30 degrees

Components of applied force.
fx = F*cos30
Fy = F*Sin30

Ff is frictional force
Ff = us*Normal Force(N) ------------------(1)

from newton second law :
sum(F) = 0, for x direction
Fx - Ff = 0
Fx = Ff ------------------(2)

for y direction :
N + Fsin30 - w = 0
N = w - Fsin30

so, eqn 1 can be written,
Ff = 0.405 * ( w - Fsin30)
Ff = 0.405 * (431 - Fsin30) -------------(3)

using eqn 3 in 2 we get :
F*Cos30 = 0.405 * 431 - 0.405*F*sin30

Solving for minimum force, F
F*Cos30 + 0.405*F*sin30 = 0.405 * 431
F( Cos30 + 0.405*sin30 ) = (0.405 * 431)
F = (0.405 * 431) / ( Cos30 + 0.405*sin30 )
F = (0.405 * 431) / ( 0.866 + 0.405*0.5)
F = 163.364 N

PART B:
If we sum moments about the point of contact between the rear legs and the floor, then the normal and frictional forces of the rear legs and the frictional force of the front legs have zero arm and thus zero moment.

The remaining forces are the driving force, the weight of the desk, the normal force of the front legs. I choose CW as positive moment .
(Ff = frictional force front legs, fr= frictional force back legs)

0 = 71F + (104/2)mg - 104Ff
0 = 0.780(163.364) + 431*1.30/2 - 1.30Ff
Ff = 313.52 N

as,
Ff = us*Nf (Nf = nornal force front legs)
Nf = 313.52/us
Nf = 313.52/0.405
Nf = 774.123 N

For back or rear legs :
using our summation of vertical forces
313.52 + Fr = 431
Fr = 117.48 N

also,
Fr = us*Nr (Nr = nornal force back legs)
Nr = 117.48/us
Nr = 117.48/0.405
Nr = 290.074 N