Forces in Cables: Elevators and Cranes
Your safety in an elevator depends on three factors – the strength of the cable, the weight hanging on the cable and the acceleration created by the motor. Let’s look at the effects of these factors in the following situations as we solve some standard types of “forces in cables” questions.
Upward Motion A mass hanging from a rope “uses up” some of the rope’s strength. When the motor tries to pull up the rope, it must exert on force on the rope and attached mass. If the motor tries to yank up the rope too quickly (too large an acceleration), it pulls with too large a force and the rope will part. Think about this: you can lift a full pop can on a hair as long as you don’t pull up too quickly.
Q. A 100 kg mass hangs from a cable that can support 4 000 N. What is the safe upward acceleration?
a) How much strength is left in the cable? Change 100 kg to N and subtract. 4 000 N – 980 = 3020 N
b) What is the safe maximum acceleration?
F = ma 3020 N = 100 kg * a or 3020 N / 100 kg = a = 30.2 m/s2
Q. A rope can support 5000 N and its motor accelerates it up at 4 m/s2. What maximum weight can the
rope support?
a) This is just the reverse type of problem and needs a bit of algebra. We use F = ma again.
5 000 N – C = C/9.8 * 4 m/s2 The C represents the total weight of what ever is hanging on the rope. The term “C/9.8” is expressing the same weight as C as a mass, in kg. Why? Because, on the right side of the equation F = ma we must show the mass of the object hanging on the rope.
b) Now, gather the similar terms by bringing the C over to the right side.
5 000 N – C = 4C/9.8 or 5 000 N = 1C + 0.41 C. But wait! If the C coming over from the left is in N but we have kg on the right side, what’s going on? Simple: recall that a mass value multiplied by an acceleration value becomes an N value. So, when we multiply the “C/9.8” term (a kg value) by the 4 m/s2 (an acceleration value) to get 4C/9.8 (or 0.41C), we are creating a N value. So, it’s OK to combine the C (in N) coming over from the left with the 0.41 C on the right side because they’re both N values.
c) Now finish. 5 000 N = 1.41 C or 5 000 N/1.41 = C = 3546.1 N
Q. A 50 kg cage hangs from a rope that can support 7 000 N. If its motor accelerates it up at 7 m/s2, what
maximum weight of cargo can be placed in the cage?
a) Change 50 kg into weight. W = mg or W = 50 kg * 9.8 m/s2 = 490 N
b) Set up the equation. See that we must use separate terms for the cage and its cargo on both sides, N on
the left and kg on the right.
7 000 N – 490 N – C = (50 kg + C/9.8) 7 m/s2
6510 N – C = (350 N + 7C/9.8)
6510 N – 350 N = 1C + 0.71 C
6160 N = 1.71 C or C = 6160 N/1.71 = 3602.34 N
Descending Motion In this type of problem, a crane is located on the 20th floor of a skyscraper under construction. When the crane’s cable is connected to some load and the boom swings out into space to lower the load, the weight alarm sounds. What to do? Remove some weight of course. How ? Not by just reaching out and picking off some material because the boom is already too far out. And, if something’s not done immediately, the overload will bend the boom, pull over the crane, break the cable or all of these. What to do? Release the brake on the cable drum and let the cable begin to accelerate toward the ground. How will this work? Think about this situation: you are hanging on a length of fishing line that can support just 50 lbs. If you hold fast to the line, it will break of course. What if you released your grip and fell to the floor? All your weight would be removed from the fishing line and so it would not break (although you might). Letting yourself fall is a way of “removing” weight from the weak line. Did you need to remove all your weight from the line by totally releasing your grip and freefalling, accelerating down at 9.8 m/s2? No, just the weight beyond the 50 lbs. You could have had a partial grip on the fishing line that would have allowed you to accelerate down at a lesser rate but quickly enough to “remove” just the required weight from the line.
Now of course with a crane at a construction site there are all sorts of issues: how did the overload occur in the first place (was no one watching the weight being attached to the crane cable?), the cost of smashing the load on the ground, the cost of a damaged crane, the cost of broken cable (if nothing is done), and the huge safety issue of things plummeting earthward.
But, how to find the correct rate of acceleration downward? Read on, physics student. It’s based on F = ma. The F is the extra weight that must be “removed” from the cable; the m is the mass of the entire load because it is the entire load that is allowed to accelerate downward to keep the cable from parting.
Q. A cable can withstand 1 000 N but is overloaded with a 1 500 N load. What should the crane operator do?
a) Find the overload. 1 500 N – 1 000 N = 500 N
b) Change the cargo weight to mass. 1 500 N/9.8 = 153.1 kg.
c) Find the downward acceleration. Use F = ma
F = ma
500 N = 153.1 kg * a
a = 500 N/153.1 kg = 3.26 m/s2
Note: allowing the load to descend at a constant rate will not do the trick. If you stand on a scale in an elevator and watch the dial you will see the reading jump only when you start or stop. Newton’s motion ideas tell us force is needed to makes things speed up, slow down or change direction. When the elevator is moving up or down at a constant speed, or are stopped, there is no net force acting on you and so your weight is its usual value. When the elevator starts down, the floor and the scale momentarily fall away from you (inertia tries to prevent you from descending with the elevator floor) and so your feet press with less force against the scale; your weight reading momentarily decreases. An instant later, your weight reading is normal. What about when the elevator is ascending at a constant speed and comes to a halt? Inertia tries to keep you moving up and so you momentarily lift off the scale bit: again, a short lived weight reduction. Of course, the faster the elevator is moving up before it stops, the more you will lift off the scale and the more your weight reading lessens for the instant.
When the elevator starts moving up, inertia tries to keep you from moving up with the elevator floor and so it momentarily pushes a bit harder into the bottom of the scale which pushes harder against your feet and so you have a sudden small weight increase. An instant later it is back to normal. What about when the elevator is moving down at a constant speed and comes to a stop? Inertia tries to keep you moving down and so you momentarily push against the scale with more force: again, a momentary weight increase. Of course, the faster the elevator is moving down before it stops, the more inertia drives your feet into the scale and the larger your weight reading increases for an instant.
What does this have to do with the crane situation? If the point is to “change” the weight of the load attached to the cable (to reduce the weight in this case), it must be allowed to fall faster and faster, it must be allowed to accelerate down. If the cable is only allowed to descend at a constant speed, even if it is very fast, the load will still have its full weight and the cable will part.
Horizontal Motion Let’s see how one mass can get two masses moving.
Q. A 4 kg mass sits on a horizontal, frictionless surface. It is attached to a 1 kg mass hanging over the
edge. Find the acceleration of the masses.
a) Find the weight of the object hanging over the edge. W = mg = 1 kg * 9.8 m/s2 = 9.8 N. Gravity tugs down on both masses but the one sitting on the table does nothing except try to press into the surface. The object hanging over the side tries to move downward but is held back because it is attached to the mass on the table. Because of the setup, the hanging mass is tugging sideways on the mass resting on the tabletop. And, with no friction, the mass on the table is able to slide sideways when it feels the sideways pull from the hanging object. So, both weights start to move at the same time and at the same rate of acceleration. The gravity force on the hanging object acts like a sideways pulling force on the object sitting on the tabletop. Since both objects move simultaneously, we must include two m’s in the equation. Use F = (m + m)a
9.8 N = (4 kg + 1 kg) a
9.8 N = 5 kg * a or a = 9.8 N/5 kg = 1.96 m/s2
Q. Now, with friction. Say the µ is 0.6. What is the acceleration then? The effect of friction is to “waste” some of the sideway pull just getting the two masses moving and so less is left to create the acceleration.
a) Find the weight of the object on the table. W = mg = 4 kg * 9.8 m/s2 = 39.2 N
b) Use µ = F/N to find the friction force.
0.6 = F/39.2 N or F = 0.6 * 39.2 = 23.52 N. Look! The friction force is larger than the sideways pull: no motion!
Q. Redo if the µ is just 0.1.
a) 0.1 = F/39.2 N or F = 0.1 * 39.2 N = 3.92 N
b) See how much sideways pulling force is left. 9.8 N – 3.92 N = 5.88 N.
c) Find the common acceleration of the two masses.
F = (m + m)a
5.88 N = (4 kg + 1 kg)a
5.88 N = 5 kg * a or a = 5.88 N/5 kg = 1.17 m/s2
Motion on a Slope This is a variation of horizontal motion
Q. A 2 kg mass sitting on a frictionless 150 ramp is attached to an 8 kg mass hanging down over the
upper end of ramp.
a) Find the weights of both masses. W = mg = 2 kg * 9.8 m/s2 = 19.6 N.
And, W = mg = 8 kg * 9.8 m/s2 = 78.4 N
b) As in the previous question, the weight of the hanging object becomes the pulling force trying to move the resting object. But, since the resting object, the 2 kg mass, is on a slope, part of its weight tries to move it downslope and this value must be subtracted from the pulling force created by the hanging object’s weight. So, find the downslope component of the 2 kg mass. Remember, use the weight.
F grav = sin 150 * 19.6 N = 5.07 N
c) Subtract the F grav from the pulling force (the weight of the hanging object) to find the final upslope pulling force.
78.4 N – 5.07 N = 73.33 N
d) Find the common acceleration of both objects.
F = (m + m)a
73.33 N = (2 kg + 8 kg)a
73.33 N = 10 kg * a or a = 73.33 N/10 kg = 7.33 m/s2.
Q. Now with friction. Say the µ is 0.4. Now, both friction and the F grav will work against the upslope pulling force created by the weight of the hanging mass and so the upslope acceleration will be less.
a) Find the N component of the weight of the object resting on the slope.
N = cos 150 * 19.6 N = 18.93 N
b) Use µ = F/N to find the friction force.
F = µ N = 0.4 * 18.93 N = 7.57 N
c) Find the balance of forces trying to move the object up and down the ramp. Both F grav and friction work against the upslope tug from the weight of the hanging object. Use Upslope tug – F grav – friction.
73.33 N – 5.07 N – 7.57 N = 60.69 N
d) Find the common acceleration of the two objects.
F = (m + m)a
60.69 N = (2 kg + 8 kg)a
60.69 N = 10 kg * a or a = 60.69 N/10 kg = 6.07 m/s2
Note: if the friction force were too large, there would of course be no motion.
Q. A 20 kg mass sitting on a frictionless 100 ramp is attached to an 6 kg mass hanging down over the
upper end of ramp.
We go through the same sequence but, because the object on the ramp is more massive than the one hanging over the top edge, the motion might be downward. It depends on the size of the F grav compared with the size of the upslope force coming from the weight of the hanging mass. Remember, the whole weight is not aimed down slope – just the F grav component. Let’s see which force wins out, the downslope gravity component or the upslope pulling force.
a) The 20 kg becomes 196 N and the 8 kg becomes 78.4 N.
b) F grav = sin 100 * 196 N = 34 N. The upslope force is still stronger. By how much? F = 78.4 N – 34 N = 44.4 N.
c) Find the common acceleration of both objects.
F = (m + m)a
44.4 N = (20 kg + 8 kg)a
44.4 = 28 kg * a or a = 44.4 N/28 kg = 1.59 m/s2. (Until I worked through this problem I had guessed the 20 kg mass would have won out. But, it is just the F grav that is key here, not the whole weight.)
Q. A 20 kg mass sitting on a frictionless 300 ramp is attached to an 6 kg mass hanging down over the
upper end of ramp. (We’ll use the same masses but increase the slope angle.)
a) 20 kg = 196 N and 8 kg = 78.4 N. The same so far.
b) F grav = sin 300 * 196 N = 98 N. The steeper slope increased the F grav! Now, which force wins out? The F grav, by 19.6 N. (Just do F grav – upslope tug or 98 N – 78.4 N = 19.6 N)
c) F = (m + m)a
19.6 N = (20 kg + 8 kg)a
19.6 N = 28 kg * a or a = 19.6 N/28 kg = 0.7 m/s2.
Motion on a Pulley Two unequal weights attached by a rope slung over a pulley adjust by the heavier descending and pulling up the lighter one. This device is called an Atwood Machine. Both objects accelerate at the same rate, the lighter one up and the heavier one down. We compare the balance of forces acting on the masses on both sides of the pulley. The math is based on F = ma.
Q. A 3 kg and a 12 kg mass are attached by a rope slung over a pulley. The 3 kg mass is on the left. Find the common acceleration rate of both masses.
a) Find the weights of both masses. W = mg = 3 kg * 9.8 m/s2 = 29.4 N. And, the 12 kg mass becomes
117.6 N
b) Compare tensions and weights with the direction in which each side moves. Of course, the left side ascends but the weight force is directed down so it is given a “–”sign. Because the tension in the rope (created by the fibers pulling up) is in the same direction in which the mass moves, it is given a “+” value. Of course, the right side descends and because the weight force on that side is directed down (the weight force is always down), it is given a “+“ sign. What about the tension on the right side? Because the tension in the rope (created by the fibers pulling up) is in the opposite direction in which the mass moves, it is given a “– ” value.
c) Now, show the balance of forces by “adding equations”.
left: F = ma
right: F = ma
left: + T – 29.4 N = 3 kg * a
right: – T + 117.6 N = 12 kg * a
0 + 88.2 N = 15 a or a = 88.2 N/15 = 5.88 m/s2