Consider a #"3 kg"# box on a flat surface. If the coefficient of static friction is #0.5#, #(a)# what is the pulling force needed to move the box, and #(b)# if one pulls at a #60^@# angle with respect to the horizontal, what force is needed instead?

Well, if you want to move the particle, you'll want the sum of the forces in the #x# direction to be greater than zero.

Furthermore, with regards to part #(b)#, pulling at #60^@# with respect to the horizontal will add a vertical component to the force, thereby requiring that the overall pulling force be higher than in part #(a)# because #-1 <= costheta <= 1# and #-1 <= sintheta <= 1#.

I got #vecF_p >= "14.72 N"# in #(a)# and #"15.77 N"# in #(b)#.

DISCLAIMER: LONG ANSWER!

First, we construct a free-body diagram for this. It is very similar to an inclined ramp, except it is horizontal:

where #vecF_g = mvecg# is the force due to gravity, #vecF_N# is the normal force (#vecF_s = mu_svecF_N#), and #vecF_s# is the static friction force.

(We assume the pulling force #vecF_p# is horizontal for now, but in #(b)# we can simply revise this diagram by rotating the #vecF_p# vector #60^@# counterclockwise.)

In this case, we can define down as positive #y# and right as positive #x# (therefore, #vecg > 0#).

#a)# We want to move the particle (which we've simply drawn as a box), so we want to overcome the sum of the horizontal forces, i.e. that sum should be positive to move rightwards. The vertical forces though can sum to #0# because perpendicular forces don't do any work.

#sum_i vecF_(i,x) = vecF_p - vecF_s >= 0# #" "bb((1))#

#sum_i vecF_(i,y) = vecF_g - vecF_N = 0# #" "bb((2))#

From #(2)# and #(1)#, respectively:

#mvecg = vecF_N#

#vecF_p >= vecF_s#

We see that the pulling force in the horizontal direction must overcome the static friction force, which makes physical sense.

So, plugging #vecF_N# into #(1)# as #mvecg#:

#=> color(green)(vecF_p >= [mu_svecF_N -= mu_smvecg])#

This means:

#vecF_p >= (0.5)("3 kg")("9.81 m/s"^2)#

#=> color(blue)(vecF_p >= "14.72 N")#

So, the force needs to be larger than #bb"14.72 N"# to move the particle, but if it is equal, then the particle will barely begin to move.

#b)# The main difference between #(b)# and #(a)# is that now we've decided to pull the particle in the direction #theta = 60^@# from the horizontal.

This adds a vertical component to the force, where #vecF_(p,x)^2 + vecF_(p,y)^2 = vecF_(p)^2# from the Pythagorean theorem.

Since #sin# and #cos# are both between #-1# and #1#, the overall force in the horizontal direction will be weakened relative to no vertical component, and thus #vecF_p# should be larger than in part #bb((a))#.

Now, we need to revise the sums:

#sum_i vecF_(i,x) = vecF_pcostheta - vecF_s >= 0# #" "bb((3))#

#sum_i vecF_(i,y) = vecF_g - vecF_N - vecF_p sintheta = 0# #" "bb((4))#

Now, as we solved #(2)# in #(a)#, we now solve #(4)# in #(b)# for #vecF_N#:

#vecF_N = mvecg - vecF_p sintheta#

Plugging #vecF_N# back into #(3)#:

#vecF_pcostheta - mu_svecF_N >= 0#

#=> vecF_pcostheta - mu_smvecg + mu_svecF_p sintheta >= 0#

We can then isolate #vecF_p#:

#vecF_pcostheta + mu_svecF_p sintheta >= mu_smvecg#

#=> vecF_p(costheta + mu_s sintheta) >= mu_smvecg#

#=> color(green)(vecF_p >= (mu_smvecg)/(costheta + mu_s sintheta))#

Therefore, the pulling force is now:

#color(blue)(vecF_p) >= (0.5 cdot "3 kg" cdot "9.81 m/s"^2)/(cos 60^@ + 0.5cdot sin60^@)#

#>=# #color(blue)("15.77 N")#

which is indeed higher than it was in part #(a)# as expected.