Question

I need help with Q9 c, I’ve done a and b if it helps, b=325 but I am just stuck on question 9C, I know the answer is 500N (as it’s in the back of the book) but I just don’t get how to get there. Many thanks in advance.

Answer 1

Step-by-step explanation:

I know you've already done parts a and b, but I'll show the work for that before I do c.

Draw two free body diagrams, one for the car and one for the trailer.  The car pulls the trailer forward with a tension force T, so the trailer pulls backward on the car with an equal and opposite force T.

The car also has a 1200 N forward force from the engine, and a 200 N backwards force from resistance.

The trailer has a backwards resistance force of 100 N.

Sum of forces on the car:

∑F = ma

1200 − 200 − T = 900a

1000 − T = 900a

Sum of forces on the trailer:

∑F = ma

T − 100 = 300a

To solve the system of equations, first add the equations together.

1000 − 100 = 1200a

900 = 1200a

a = 0.75 m/s²

Plug back into either equation to find the tension force:

T = 325 N

Now for part c, draw new free body diagrams for the car and trailer.  This time, the car is pushing back on the trailer to slow it down.  So the trailer is pushing forward on the car with an equal and opposite force.  The magnitude of that tension force is given to be 100 N.

The car also has a backwards 200 N force from resistance, and a backwards brake force F.

The trailer has a backwards 100 N force from resistance.

Sum of forces on the car:

∑F = ma

100 − 200 − F = 900a

-100 − F = 900a

Sum of forces on the trailer:

∑F = ma

-100 − 100 = 300a

-200 = 300a

a = -⅔

Plugging into the first equation:

-100 − F = 900 (-⅔)

-100 − F = -600

F = 500 N