Sample Force Problems.....................

Example 1: A balloon floats a 100 kg gondola upward with acceleration 1.2 (m/s2). What force is on the gondola? (this problem is similar to many you have done)

Draw the above diagram (a "free body" diagram) first. Then write Newton's 2nd Law:

FnetY   =    ma

F(up) - W(down) = ma

F(up) = ma + W(down)

F(up) = ma + mg

F(up) = (100 kg)(1.2 m/s^2) + (100 kg)(9.8 m/s^2)

F(up) = 120 N + 980 N

F(up) = 1100 N ................QED



Equilibrium means that the forces are all balanced...or, in other words, they cancel each other out such that the object remains stationary, or STATIC:

SFX = 0   and  SFY = 0

This equation expresses that fact.

For the 2nd example, we'll look at a 120 kilogram traffic light hanging from a couple of cables. We want to find the tension in each cable....

We first draw the "free-body" diagram for the object (traffic light):

A "free-body" diagram is: a drawing of the object alone with the forces acting on it labeled: it is VERY important that the forces shown are ONLY the forces acting directly on the object. For instance, we don't care about the forces holding the cables in place at the other end.

Now the second step is to write down Newton's 2nd Law in both x and y directions (for the equilibrium case):

Hmmm. This means we just can't use T1 and T2 as they are drawn. We have to figure out the forces acting in the x-direction and the y-direction (hey, nobody promised this would be solved in two steps):

The above diagram shows how we can break up T1 and T2 into x and y components. The WEIGHT of the traffic light, (mass)(gravity), or mg, is already pointing in the -y direction so it needs no further breakdown.

Now we're ready to plug the force variables into the equations. Leaving the right side of the equations alone, we SUM the forces on the LEFT side of the equation and then solve for the unknowns (remember to note + and - directions!!!! for example, Tx1 will be added as -Tx1....why?)...........

OK, so now you can hang traffic lights and pictures...just don't hang any around my house...


Example 3: friction

A 1000 kg car moving at 10 m/s hits the brakes, the wheels lock, and he skids for 100 meters. What is the coefficient of friction for the ice/tire combo?

Again, we draw the above diagram, and then (and only then!) write Newton's 2nd Law (noting that the normal force, F(n) is equal to the weight W:

SFX = ma

-f(k) = ma

-m(k) x F(n) = ma

-m(k) x (mass)(gravity) = ma

m(k) = - ma/mg

m(k) = - a/g

So we need the acceleration!! Which we obtain from those equations of constant acceleration:

v(f)^2 = v(i)^2 + 2a(Dd)

a = {v(f)^2 - v(i)^2}/{2(Dd)}

a = {0 - 100 m^2/s^2)/(200 m)

a = - 0.5 m/s^2

Now we can finally find m(k):

m(k) = - (- 0.5 m/s^2)/(9.8 m/s^2)

m(k) = 0.05....................QED