PLAIN AND CIVIL: EXAMPLE 7.0.4: EQUILIBRIUM OF COPLANAR FORCES

Friday, 8 March 2019

EXAMPLE 7.0.4: EQUILIBRIUM OF COPLANAR FORCES

A 2400-lb tractor is used to lift 900lb of gravel. Determine the reaction at each of the two (a) rear wheels at A, (b) front wheels at B.

Analyze the system by using FBD.

Using the FBD, besides the summation of forces - horizontal and vertical, use the third equation of equilibrium in this case.

   $\sum F_{X}=0$

   $\sum F_{Y}=0$

   $\sum M=0$

With the given loads projected vertically, there will not be any horizontal reaction.

If we use summation of vertical loads, the equation will end up with two unknowns

$R_{A}+R_{B}-2400-900=0$

   $R_{A}+R_{B}=3300lb$

The way to use summation of moments is to position it in a way which cancels the other unknown. For example, we start with using summation of moments to solve for Ra, the point to be considered should be point B. By taking moments at point B zeroes out Rb because there is no perpendicular distance of Rb from point B.

$\sum M_{B}=0$ clockwise-positive

$R_{A}\left ( 60in \right )-2400lb\left ( 40in \right )+900lb\left ( 50in \right )=0$

${\color{Red} \mathbf{R_{A}=850lb}}$

From here, you can use $\sum F_{Y}=0$ or $\sum M=0$ about point A, to solve for the other reaction Rb.

By using $\sum F_{Y}=0$ :

   $850+R_{B}=3300$

   ${\color{Red} \mathbf{R_{B}=2450lb}}$

By using $\sum M_{A}=0$

$2400lb\left ( 20in \right )-R_{B}\left ( 60in \right )+900lb\left ( 110in \right )=0$

   ${\color{Red} \mathbf{R_{B}=2450lb}}$

Both answers are for (a) two rear wheels and (b) two front wheels, thus

${\color{Red} \boldsymbol{R_{A}=\frac{850}{2}=425lb}}$
${\color{Red} \boldsymbol{R_{B}=\frac{2450}{2}=1225lb}}$

RELATED TOPICS

EQUILIBRIUM OF FORCE SYSTEMS