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

Temka [501]

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 $F_{s} < F$

8 0

3 years ago

Gymnasts often practice on foam floors, which increase the collision time when a gymnast falls. What effect does this have on co

krok68 [10]

Gymnasts often practice on foam floors, which increase the collision time when a gymnast falls. What effect does this have on collisions

The effect of the gymnast on the collision will increase. An elastic collision is when two bodies collide and separates after collision conserving the total kinetic energy before and after collision.

8 0

3 years ago

Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible fricti

Bas_tet [7]

Answer:

v = 0.059 m/s

Explanation:

To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:

$mv_{1i}+Mv_{2i}=(m+M)v$ (1)