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a)

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To find the resultant force in a certain direction when there is more than one force acting on an object you can resolve the forces. You resolve in the direction of acceleration. Using the letter $R$, with an arrow will indicate the direction the forces are being resolved in, for example $R(\\rightarrow)$.

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In this case $R(\\rightarrow)$ means apply $F=ma$ in the positive direction. So we need to resolve the equation 

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\\begin{align}F & = ma \\\\
                   C - \\var{FA} & = \\var{K} \\times \\var{a} \\\\
                                  C & = \\var{K*a} + \\var{FA} \\\\
                                       & = \\var{K*a + FA}. \\end{align}

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Here the forces, $F=C - \\var{FA}$ because $\\var{FA}$ is acting in the opposite direction to $C$. The mass is $m=\\var{K}$ and the acceleration is $a=\\var{a}$. Therefore $C = \\var{K*a+FA}\\  \\mathrm{N}$.

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b)

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In this case we resolve in the direction perpendicular to acceleration so we use $R(\\uparrow)$ which means apply $F=ma$ in the vertical direction, where acceleration, $a=0$ because there is no vertical acceration.

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\\begin{align} F & = ma \\\\
                    B - \\var{FD}g & = \\var{K} \\times 0 \\\\
                                    B & = 0 + \\left(\\var{FD} \\times 9.8\\right) \\\\
                                       & = \\var{FD*9.8} \\end{align} 
Here the forces, $F=B - \\var{FD}g$ because $\\var{FD}g$ is acting downwards which is the opposite direction to $B$. Therefore $B = \\var{FD*9.8} \\ \\mathrm{N}$.

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c)

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It is easier to take the positive direction as the direction of the acceleration. Therefore we use $R(\\leftarrow)$ because we are decelerating. 

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\\begin{align} F & = ma \\\\
                    \\var{FA2} - C & = \\var{K2} \\times \\var{a2} \\\\
                           C & = \\var{FA2} - \\left( \\var{K2} \\times \\var{a2} \\right) \\\\
                              & = \\var{FA2 - K2*a2}. \\end{align}

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Here the forces, $F = \\var{FA2} - C$ because $\\var{FA2}$ is in the direction of acceleration and $C$ is acting in the opposite direction. The mass is $m = \\var{K2}$ and $a=\\var{a2}$. Therefore $C = \\var{FA2 - K2*a2} \\ \\mathrm{N}$.

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", "rulesets": {}, "parts": [{"prompt": "

Suppose that the acceleration $a = \\var{a}$ and the mass of the particle $K = \\var{K} \\ \\mathrm{kg}$. Find the force $C \\ \\mathrm{N}$ if $A = \\var{FA}\\ \\mathrm{N}$.

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$C = $ [[0]] $\\mathrm{N}$

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You have not given your answer to the correct precision.

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Find the magnitude of the force $B$ if $D =\\simplify{{FD}g} \\ \\mathrm{N}$.

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Consider another particle which has mass $\\var{K2} \\ \\mathrm{kg}$ and is now decelerating horizontally at $\\var{a2}\\ \\mathrm{ms^{-2}}$. If $A = \\var{FA2}\\ \\mathrm{N}$ what is $C$?

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Consider the following diagram of a particle of mass $K \\ \\mathrm{kg}$, moving left-to-right. The forces $A, B, C$ and $D$ are acting upon it, producing a horizontal acceleration of $a \\, \\mathrm{ms^{-2}}$ 

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The acceleration due to gravity is $9.8 \\, \\mathrm{ms}^{-2}$. Answer all the following questions to 3 decimal places.

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acceleration

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Force A

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mass

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mass 2

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downforce D 2

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Force D

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downforce for part d)

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acceleration

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Using $F=ma$ to find magnitudes of unknown forces acting on an accelerating body.

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