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Chandler J.
Engineering Student at EWU
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Mechanical Engineering
TutorMe
Question:

A box weighing 13kg sits on the ground. If the coefficient of static friction between the box and the ground is 0.7, what is the minimum force that must be applied to get the box to move?

Chandler J.

The answer to this question is found by using a simple free-body diagram. The box has four forces acting on it: 1) The weight of the box acting straight down: $$W=13*(9.81)=127.5 N$$ 2) The normal force from the ground acting upward (equal to the weight) $$F_{N}=127.5N$$ 3) The frictional force acting opposite the direction of the applied force 4) The applied force In order to calculate the friction force, we use the equation: $$F_{fric}=\mu*F_{N}$$ Plug in what we know: $$\mu=0.7$$ $$F_{N}=127.5 N$$ $$F_{fric}= (0.7)*(127.5)=89.25 N$$ Since the vertical forces are equal and opposite, they cancel each other out so only the horizontal forces remain. We would need to apply a minimum force of 89.25 N to overcome the 89.25 N frictional force.

Algebra
TutorMe
Question:

Solve the system of equations for x and y: $$5x-12y=36$$ $$x+2y=5$$

Chandler J.

To answer this one we need to evaluate at each equation individually first. We can solve the top equation for x in terms of y, then substitute what we got in the top equation directly into the bottom equation. $$5x - 12y=36$$ ...moving y to the other side becomes $$5x=36+12y$$ ...simplifying further $$x=\frac{36+12y}{5}$$ Now we have x in terms of y. Substituting x into the bottom equation gives $$(\frac{36+12y}{5})+2y=5$$ We can make this easier by multiplying both sides by 5 $$5*((\frac{36+12y}{5})+2y)=(5)*5$$ $$({36+12y})+10y=25$$ Now we have $$36+22y=25$$ solving for y... $$y=-0.5$$ We can now plug in y to any of the equations above to solve for x $$x+2y=5$$ $$x+2(-0.5)=5$$ $$x=6$$

Physics
TutorMe
Question:

A bullet is fired from the ground with an incline of 35 degrees above the horizontal. If the bullet's initial velocity is 165 m/s, how high did the bullet go? (Ignore air resistance)

Chandler J.

The bullet was fired at an angle, so we need to find the vertical and horizontal components of the bullet's trajectory. However the question is asking how high the bullet traveled, so we only care about the vertical component. We find the vertical component using trigonometry: $$165*sin(35)=94.64$$ This means that if we ignored the horizontal movement, the bullet would have an initial velocity 94.64 m/s straight up into the air. Now that we know the bullet's initial vertical velocity, we can use the equation from projectile motion to solve for the bullet's vertical displacement. $$(V_{f})^{2}=(V_{o}) ^{2}+2a(x_{f}-x_{o})$$ Plug in what we know: $$V_{f}=0$$ because the bullet will not be moving at the top of it's trajectory $$V_{o}=94.64$$ because that is the bullet's initial vertical speed $$a=-9.81$$ because that is the effect from gravity $$x_{o}=0$$ because we are assuming that the bullet was fired from the ground Now we solve for $$x_{f}$$: $$(0)^{2}=(94.64)^{2}+2(-9.81)(x_{f}-0)$$ $$x_{f}=456.5m$$ The bullet achieved a maximum height of 456.5 meters.

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