3. A rugby lployer sits on a scrum machine that weighs F. 2004. Given that the cooffiaut of static friction is 0.18 mathrm(lls)= the coefficient of kinetic friction is 0.15 and the force requirod to push both the scrum machine and the players into motion is 81 mathrm(~N)

To solve this problem, we need to determine the force required to overcome static friction and set the scrum machine and players in motion. The force of static friction can be calculated using the formula:<br /><br />\[ F_{\text{friction}} = \mu_s \cdot N \]<br /><br />where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force. In this case, the normal force is equal to the weight of the scrum machine, which is \( 2004 \, \text{N} \).<br /><br />\[ F_{\text{friction}} = 0.18 \cdot 2004 \, \text{N} = 361.28 \, \text{N} \]<br /><br />Since the force required to push both the scrum machine and the players into motion is \( 81 \, \text{N} \), it means that this force is equal to the force of kinetic friction once the motion has started. The force of kinetic friction can be calculated using the formula:<br /><br />\[ F_{\text{kinetic friction}} = \mu_k \cdot N \]<br /><br />where \( \mu_k \) is the coefficient of kinetic friction. In this case, the normal force is still equal to the weight of the scrum machine, which is \( 2004 \, \text{N} \).<br /><br />\[ F_{\text{kinetic friction}} = 0.15 \cdot 2004 \, \text{N} = 301.6 \, \text{N} \]<br /><br />Therefore, the force required to push both the scrum machine and the players into motion is \( 81 \, \text{N} \), which is equal to the force of kinetic friction once the motion has started.