Question

A farm tractor tows a 3300 kg trailer up a 14 degree incline with a steady speed of 2.8 m/s. What force does the tractor exert on the trailer?

Answer 1

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:

$\begin{gathered} F=\text{ force of the tractor} \\ m=\text{ mass of the trailer} \\ g=\text{ acceleratio of gravity} \end{gathered}$

Now, we add the forces in the direction of the incline:

$\Sigma F_x=F-mg_x$

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":

$\sin14=\frac{mg_x}{mg}$

Now, we multiply both sides by "mg":

$mg\sin14=mg_x$

Now, we substitute the values of "m" and "g":

$(3300kg)(9.8\frac{m}{s^2})\sin14=mg_x$

Solving the operations:

$7823.75N=mg_x$

Now, we substitute the value in the sum of forces:

$\Sigma F_x=F-7823.75N$

Since the object is moving at a steady speed this means that the sum of forces is zero:

$F-7823.75N=0$

Now, we add 7823.75N to both sides:

$F=7823.75N$

Therefore, the tractor exerts a force of 7823.75N