We are asked to determine the force that the tractors is exerting on a trailer up an incline. A free-body diagram of the situation is the following:
Where:
[tex]\begin{gathered} F=\text{ force of the tractor} \\ m=\text{ mass of the trailer} \\ g=\text{ acceleratio of gravity} \end{gathered}[/tex]Now, we add the forces in the direction of the incline:
[tex]\Sigma F_x=F-mg_x[/tex]To determine the x-component of "mg" we use the following right triangle:
Now, we use the function sine to determine the value of "mgx":
[tex]\sin14=\frac{mg_x}{mg}[/tex]Now, we multiply both sides by "mg":
[tex]mg\sin14=mg_x[/tex]Now, we substitute the values of "m" and "g":
[tex](3300kg)(9.8\frac{m}{s^2})\sin14=mg_x[/tex]Solving the operations:
[tex]7823.75N=mg_x[/tex]Now, we substitute the value in the sum of forces:
[tex]\Sigma F_x=F-7823.75N[/tex]Since the object is moving at a steady speed this means that the sum of forces is zero:
[tex]F-7823.75N=0[/tex]Now, we add 7823.75N to both sides:
[tex]F=7823.75N[/tex]Therefore, the tractor exerts a force of 7823.75N
