# Net Force Problems.

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Net Force Problems

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

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)

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

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

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.

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

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

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

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