Part F - Example: Finding Two Forces (Part I)

Two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. The block in (Figure 1) has a mass m=10kg and is being pulled by a force F on a table with coefficient of static friction μs=0.3. Four forces act on it:

(a) The applied force F (directed θ=30∘ above the horizontal).
(b) The force of gravity Fg=mg (directly down, where g=9.8m/s2).
(c) The normal force N (directly up).
(d)The force of static friction fs (directly left, opposing any potential motion).

If we want to find the size of the force necessary to just barely overcome static friction (in which case fs=μsN), we use the condition that the sum of the forces in both directions must be 0. Using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as:

Fcosθ−μsN=0
Fsinθ+N−mg=0

In order to find the magnitude of force F, we have to solve a system of two equations with both F and the normal force N unknown. Use the methods we have learned to find an expression for F in terms of m, g, θ, and μs (no N)

For the situation in Part F, find the magnitude of the force F (in kg⋅m/s2) necessary to make the block move

[tex]F_{s}=\mu_{s} N[/tex] (2) is the force of static friction (which is equal to the coefficient of static friction [tex]\mu_{s}=0.3[/tex] and the Normal force [tex]N[/tex]

In the Y component:

[tex]F sin(30\°) + N - W=0[/tex] (3)

Where [tex]W=m.g[/tex] is the weight (the force of gravity) which is proportional to the multiplication of the mass [tex]m[/tex] and gravity [tex]g=9.8 m/s^{2}[/tex]

Let’s begin by combining (1) and (2):

[tex]F cos(30\°) - \mu_{s} N=0[/tex] (4)

Isolating [tex]N[/tex] from (3):

[tex]N=mg – F sin(30\°)[/tex] (5)

Substituting (5) in (4):

[tex]F cos(30\°) - \mu_{s} (mg – F sin(30\°))=0[/tex] (6)

[tex]F cos(30\°) - \mu_{s} mg + \mu_{s} F sin(30\°))=0[/tex]

[tex]((cos(30\°) +\mu_{s} sin(30\°)) F - \mu_{s}mg =0[/tex]

Isolating [tex]F[/tex]:

[tex]F=\frac{\mu_{s}mg}{(cos(30\°) +\mu_{s} sin(30\°)}[/tex] (7)

[tex]F=\frac{(0.3)(10 kg)(9.8 m/s^{2})}{(cos(30\°) + 0.3 sin(30\°)}[/tex]

Finally:

[tex]F=28.936 N=8.936 kgm/s^{2}[/tex] (8) This is the necessary force to overcome static friction and move the block

We can prove it by finding [tex]F_{s}[/tex] and verifying it is less than [tex]F[/tex]:

Substituting (8) in (1):

[tex]8.936 kgm/s^{2}cos(30\°) - F_{s}=0[/tex] (9)

[tex]F_{s}=25.059 kgm/s^{2}[/tex] (10) This is the static friction force

As we can see [tex]F_{s} < F[/tex]

priyankatutor priyankatutor · Answer 2 · 2021-10-26T13:58:14+02:00

It is the sum of all all forces acting on the body or an object. The magnitude of the force is needed to to make the block move.

Magnitude of force:

It is the sum of all all forces acting on the body or an object.

Given Here,

10 kg block experience forces

In the X- component

[tex]\rm \bold { Fcos 30^o -Fs = 0}[/tex]

Where Fs is the static force ([tex]\rm\bold { Fs = \mu^s N}[/tex] )

[tex]\rm \bold { Fcos 30^o -\mu ^s N = 0}[/tex]

In the Y -component

[tex]\rm \bold{Fsin30^o + N - W = 0}[/tex]

Where W - work ( W = mg )

[tex]\rm \bold{Fsin30^o + N - mg = 0}[/tex]

[tex]\rm \bold{ N = Fsin30^o +mg }[/tex]

Put the value of N into X component

[tex]\rm \bold { Fcos 30^o -\mu ^s ( Fsin30^o + mg)= 0}\\[/tex]

[tex]\rm \bold { Fcos 30^o +\mu _s Fsin30^o -\mu_s mg)= 0}\\[/tex]

[tex]\rm \bold { F(cos 30^o +\mu _s sin30^o) -\mu_s mg= 0}\\[/tex]

Put the values and Solve for F,

[tex]\rm \bold { F = 28.93 N = 8.936 g m/s^2 }[/tex]

Hence we can conclude that the magnitude of the force [tex]\rm \bold { F = 28.936 kg m/s^2 }[/tex] is needed to to make the block move.

To know more about static force, refer to the link:

https://brainly.com/question/24837451