1. Net Force Problems

2. 1. An airplane has 4 forces acting on it: weight, drag (air friction), the thrust of the engines and the lift force caused by the flow of air over the wings.
55,000 N
5000 N up
If the pilot wishes the airplane to move forward, then the ______ must be greater than the _______.
Flift
Fdrag
20,000 N
Fthrust
20,000 N
Airplane
Fweight
50,000 N
thrust
drag

3. 2. A 200 N table has a 30 N book resting upon it.
Ftable
Book
Fweight of book
30 N = equal and opposite force of the table
pushing up on the book (called the normal
force)
Book
30 N = mass of book x acceleration (9.81 m/s2)

4. Each rope must exert a force of 75 N. 150 N = 75 N + 75 N
3. A child weighing 150 N is sitting on a swing. The swing is equally supported by 2 ropes, one on each side. What is the force exerted by each rope?
Each rope must exert a force of 75 N. 150 N = 75 N + 75 N
Frope
Frope = 75 N, which is the tension force
Swing seat
Fchild = 150 N, which is the force of
weight due to gravity

5. Fchain 2 = Fweight - Fchain 1 = 1500 N – 600 N = 900 N
4. Two chains are used to support a small boat weighing 1500 N. One chain has a tension of 600 N. What is the force exerted by the other chain?
Fchain 2 = Fweight - Fchain 1
= 1500 N – 600 N
= 900 N
Fchain 2
Fchain 1 = 600 N
Boat
Fweight = 1500 N

6. 5. A heavy box weighing 1000 N sits on the floor
5. A heavy box weighing 1000 N sits on the floor. You press down on the box with a force of 450 N. What is the normal force on the box? (Same method as #4)
Fn = 1450 N
Box
Fp = 450 N
Fw = 1000 N
A normal force is perpendicular to the plane of action.

7. 6. A 40 N cat stands on a chair. If the normal force on each of the cat’s back feet is 12 N, what is the normal force on each front foot? (Same method as #4)
The normal force must equal the force of the cat’s weight 40 N = 2(12 N) + 2x 40 N = 24 N + 2x 16 N = 2x 8 N = x Therefore, the normal force on each front foot is 8 N

8. 7. Two forces are acting on a pencil, pushing it to the right
7. Two forces are acting on a pencil, pushing it to the right. One force is 9 N and the other is 6 N. There is a 6 N force pushing the pencil to the left. What is the net force acting on the pencil? (Don’t forget the direction!)
Fp = 9 N
Fp = 6 N
Pencil
Fp = 6 N
The net force acting on the pencil is 9 Newtons to the right.
Fp = 9 N
Pencil

9. 8. A train is climbing a gradual hill
8. A train is climbing a gradual hill. The weight of the train creates a downhill force of 150,000 N. Friction creates an additional force of 25,000 N acting in the same direction. How much force does the train’s engine need to produce so that the train is in equilibrium (not sliding back down the hill)?
Ft = ? N
Train
Ff = 25,000 N
Fw = 150,000 N
Ft = Ff + Fw Ft = 25,000 N + 150,000 N = 175,000 N