Question

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

Answer 1

Answer: $F=28.936 kg/m s^{2}$

Explanation:

According to the given information (and figure attached), the block with mass $m=10 kg$ has the following forces acting on it:

In the X component:

$F cos(30\°) - F_{s}=0$ (1)

Where:

$F$ is the applied force directed $30\°$ above the horizontal

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

In the Y component:

$F sin(30\°) + N - W=0$ (3)

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

Let’s begin by combining (1) and (2):

$F cos(30\°) - \mu_{s} N=0$ (4)

Isolating $N$ from (3):

$N=mg – F sin(30\°)$ (5)

Substituting (5) in (4):

$F cos(30\°) - \mu_{s} (mg – F sin(30\°))=0$ (6)

$F cos(30\°) - \mu_{s} mg + \mu_{s} F sin(30\°))=0$  

$((cos(30\°) +\mu_{s} sin(30\°)) F - \mu_{s}mg =0$  

Isolating $F$:

$F=\frac{\mu_{s}mg}{(cos(30\°) +\mu_{s} sin(30\°)}$ (7)

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

Finally:

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

We can prove it by finding $F_{s}$ and verifying it is less than $F$:

Substituting (8) in (1):

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

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

As we can see $F_{s} < F$