As I was driving home yesterday, a pickup truck pulled out in front of me with a

Question

As I was driving home yesterday, a pickup truck pulled out in front of me with a motorcycle lying down in the back. The gate of the truck was down so the only things holding the motorcycle on the truck was the (rather unhappy looking) guy in the back and the friction between the motorcycle and the metal truck bed. (By the way, this IS a true story.) Let's assume the mass of the motorcyle is 250 kg, the maximum force the guy in the back is willing to apply to the motocycle is 60 N, and the largest possible force of friction between the truck and the motorcycle is 0.10 times the normal force the truck is applying to the motorcycle. What is the maximum possible forward acceleration for the pickup that would still keep the motorcycle in the truck? Consider only the case where the road is level (horizontal), although they did need to go up and down a few small hills. Give your answer in m/s2 to three significant digits.

Explanation / Answer

I think this question is supposed to be tricky... because it seems to imply that the guy in the truck is pulling horizontally on the motorcycle, but he doesn't have to. In fact, we get a greater possible acceleration if he pulls forward and pushes down on the motorcycle at the same time.

So here's what I'll do:

First, simple case: assume the guy pulls only horizontally.

So: total horizontal force = force from guy + static friction = 65 + 0.12 * 175 * 9.80 =

= 270.8 Newtons

max acceleration = force / mass = 270.8 / 175 = 1.55 m/s^2

Now... complicated case: assume the guy pushes down on the motorcycle while also pulling forward so that his total force is F = 65 N at an angle of @ below the horizontal.

Then we have:

horizontal forces: F cos@ + 0.12 * normal = ma

vertical forces: normal - F sin@ - mg = 0

Eliminate normal by substitution: normal = F sin@ + mg so

F cos@ + 0.12 (F sin@ + mg) = ma

solve for a

a = (F/m) cos@ + 0.12 (F/m) sin@ + 0.12g

there are two unknowns... angle @ and acceleration. But we want the max possible acceleration. There is a certain angle that gives the max possible acceleration... and to find that angle we simply have to take the derivative of a with respect to @ and set it to zero.

first plug in some numbers

a = (65/175) cos@ + 0.12 (65/175) sin@ + 0.012* 9.80

or

a = 0.37143 cos@ + 0.04457 sin@ + 1.176

and the derivative

0 = -0.37143 sin@ + 0.04457 cos@   

or

tan@ = 0.04457 / 0.37143 @ = 6.843

and then the max possible acceleration is

a = 0.37143 cos6.843 + 0.04457 sin6.843 + 1.176 =

= 1.55 m/s^2 wow... ok, so it comes out almost exactly the same.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.