Question

A toy chest and its contents have a combined weight of W = 190 N. The coefficient of static friction between toy chest and floor µs is 0.450. The child in Fig. 6-35attempts to move the chest across the floor by pulling on an attached rope. (a) If ? is 46.0°, what is the magnitude of the force that the child must exert on the ropeto put the chest on the verge of moving? Determine (b) the value of ? for which F is a minimum and (c) that minimum magnitude.

Answer 1

Because we will eventually need to calculate the angle for the minimum force, we will keep our equations as symbols for now.

The force required to pull the block will be the normal force (F_n) * coefficient of friction (mu)

so pulling force = (F_n)*mu right before it moves

F_n = force of gravity - force pulling up
force of gravity = m*g
force pulling up = F * sin (angle) ---- because the sin gives us the opposite side of the triangle which is the vertical side

F_n = mg - F sin (angle)

Now to calculate the pulling force (F_p), we want the horizontal component of the rope
F_p = F cos (angle)

So, putting all of this together

F cos(angle) = mu * (mg - F sin(angle))

F cos(angle) = mu*mg - mu* F sin(angle)
----
Solve part a)
F cos(46 deg) = .450*(190 Newtons) - .450 * F sin(46 deg)

F*( (cos(46)+.45 sin(46)) = .450*190

F = 84 Newtons

part b)
What is the most efficient angle for F?
so to get this, we want cos(angle) + .45*sin(angle) to be as large as possible
You can graph cos(X)+.45*sin(X) on your calculator to find the max value or do this algebraically (please comment if you need help with this step)

for part c)
Once you find the maximum angle, plug that back into the equation and calculate the new F