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Newton’s law: F = m a

Forces cause motion. But if the forces balance each other out, the body acted upon by the forces is in equilibrium. 

The net force is zero. The object is therefore at rest or moving at a constant velocity.

(same direction and same speed at any time). That is the acceleration is zero.

 The study of body in equilibrium is called statics.

A load supported by a crane (see image) is in equilibrium between its weight and the tension in the cable.

2 forces act on the load: the tension in the string pulling up and the weight pulling down

free-body diagram

To simplify the work of a physicist or engineer, a  “free-body diagram” of the body (load here) is drawn. This is a sketch in which we draw only the object in equilibrium (some time as a point) and the forces that act on it.

EXERCICES:

Try not to look at the hints. Follow the steps: i) sketch the situation  ii) trace x,y coordinate system and attach all the forces (vectors) to the origin/
The origin is at the center of mass of the object studied. In the above example, the origin is the load.

 

1)Can you draw a body-diagram for each of the situations. The diagram must show all the forces acting on the object. The forces will include :

W (weight pulling down), N (normal force exerted by the surface on the object. Usually same size as W but opposite direction), P (pulling force or engine force, moves the object forward), T (recoil force in a spring/string, opposes the stretching of the spring/string, same size as the weight/stress but opposite direction).

A) A book standing on a table.
Hint: forces are: Normal  (up) and weight (down)
B) A car moving at a constant speed in a straight line.
Hint: forces are engine (right), friction (left) , weight (down), normal (up)
C) A sled pulled by a child at a constant speed along a flat surface.
The rope makes a 30 degrees angle with the horizontal.
Hint: forces are pulling force at an angle of 30 with horizontal, friction (left), weigh t(down), normal (up)
E) A load on a vertical spring.
Hint: tension (recoil force up), weight (down)
F) A boat at rest.
Hint: forces are weight (down), normal (Archimedes force, up)

G) a sign hanged by 2 ropes: Hint: tension along each rope, weight

H ) A skier moving down hill Hint: weight, normal perpendicular to surface along the skier body, friction along surface but opposes motion

EQUILIBRIUM IN ONE DIMENSION

 We have F+ = F-

F+ = sum of all the forces acting in one direction (called the positive direction)
F- = sum of all the forces acting in the opposite direction (called the negative direction).

 2)

A) Four persons are having a tug-of-war with a rope. Harry and Mary are on the left; Bill and Jill are on the right. Mary pulls with a force of 105 lb, Harry pulls with a force of 255lb, and Jill pulls with a force of 165 lb. With what force must Bill pull to produce equilibrium ? Sketch a free-body diagram.

 B) Find the force F that will produce equilibrium in each free-body diagram.

 

3) A front-wheel drive car is traveling at CONSTANT VELOCITY. (forces balance each others). The forces acting on the car are shown in the diagram above. Q is the force of the air on the moving car. P is the total upward force on both front wheels. A) Find the magnitude of vector P Hint: along the vertical the forces pointing up = forces pointing down. B) Find the magnitude of vector Q. Hint: along the vertical forces to the right = forces to the right C) Calculate the mass of the car. D) The 400N driving force is suddenly doubled. (Q stays constant) i) Calculate the resultant force driving the car forward. ii) Calculate the acceleration of the car Hint: F = m a . F is the force acting on an object, m is the mass, a is the acceleration.

EQUILIBRIUM IN TWO DIMENSIONS

 In an object is in equilibrium in two dimensions, the net force acting on it must be zero. For the net force to be zero, the sum of the x-components must be zero, and the sum of the y-components must be zero.

 For forces A, B and C with x-components Ax, Bx, Cx respectively, and with y-components Ay, By, and Cy, respectively, to be in equilibrium:

 The sum of x-components = 0 ; that is Ax+Bx+Cx = 0 and
The sum of the y-components =0; that is Ay+By+Cy=0

 In general, to solve equilibrium problems:

1– Draw a free-body diagram from the point at which the unknown forces act

2– Find the x-component and y-component of each force.

3. Substitute the components in the equation:

                     Sum of x-components = 0

                     Sum of y-components = 0

4– Solve of the unknowns

4) Find the forces F1 and F2 necessary to produce equilibrium in the free body-diagram shown: Hints: (try without) A)First Find the x-component of each force: Tx = _____________  (use cosine. Tx is negative) F1x= F1 F2x = 0 B) Find the y-component of each force Ty = ______________ (use sine) F1y = 0 F2y = F2 C) at equilibrium the sum of the x-components = 0 (Tx + F1x + F2x = 0) and the sum of the y-component = 0 (Ty + F1y + F2y = 0) Write the 2 equations and solve for F1 and F2 . Don’t forget the direction. If a x-component points to the left, it is negative.  If a y-component points down, it is negative.

4) You are pulling a brick (10kg) with a spring scale. The spring is moving at a constant speed. 
The spring scales reads 4N. Try without hints.
A) Sketch a free-body diagram.
hint: the pulling force is 4N and points to the right. The frictional force points to the left.
The weight points down and the normal force points up. All the forces meet at the origin (the mass of the brick).
B) find the force of friction.
Hint: sum of the x-components is 0. Tx + Fx = 0. T is the pulling force and F is the friction. Fx <0
C) find the elastic recoil of the tabletop (that is the normal force)
Hint: sum of the y-components is zero. N + W = 0 . N is the normal force and W is the weight. W <0.

5) 

5) You are pulling a block with a cord. Constant speed. The block is 2.5 kg. The cord makes an angle of 20 degrees with the horizontal and has a tension of 7N in it (T = 7N@ 20) . Find the force of friction. The elastic recoil of the tabletop (normal force). Hints: A) Make a free body diagram. It should include the tension T, the frictional force F , the normal N and the weight W B) Find the x-component and the y-component of each force acting on the block C) Write the 2 equations : sum of the x-components = 0 and sum of the y-component =0 Solve for F and N. Make sure F is negative (to the left) and W is negative (down).

6) An object in equilibrium has three forces exerted on it. A 33N force acts at 90 degrees from the x-axis 
(F1 = 33N  @ 90) and a 44N force acts at 60 degrees (F2 = 44N@ 60). 
What are the magnitude and direction of the third force F3 . Try without hint.

Hint:A) make a free-body diagram. F1 points up, F2 is in the 1st quadrant
B) Find the x-component and the y-component of F1 and F2.
F1x = 0 , F1y = F1 , F2x = _________(use cosine)  F2y = ____________ (use sine)
C) write the 2 equations for equilibrium:
F1x + F2x + F3x = 0
F1y + F2y + F3y = 0
Solve for F3x and F3Y
D) once you have F3x and F3Y it is easy to find the magnitude of F3 (Pythagorean) . Trace F3x then F3y then connect the tail of F3x to the head of F3y to find F3. Use Pythagorean theorem to find the magnitude. (F3x² +F3y² = F3²).
Use tangent to find the direction.
F3 = ___N@ _______

 

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7) A street lamp weighs 150N. It is supported by 2 wires that form an angle of 120 degrees with each other. The tensions in the wires are equal . A) what is the tension in each wire ? Hints: sketch a free-body diagram. It should include the weight down, the tension in each wire.  They meet at a concurrent point. Find the y-components of the tensions (positive, use sin (30) )  and find the y-component of the weight (negative, it points down) . Use Ty + Ty + W = 0 to solve for T . B) If the angle between the wires is reduced to 90 degrees, what is the tension in each wire ?

8) A block is at rest on an inclined plane. On the diagram: Label for forces acting on the block: - Weight - Normal force (recoil of the plane) - Frictional force

LAB: frictional force and coefficient of friction (use firefox)

9) extra for Physics major:

A 215N box is placed on an inclined plane that  makes a 35 angle with the horizontal. Find the component of the weight force parallel to the plane’s surface.

 10) extra for Physics major:

You pull your 18kg suitcase at constant speed on a horizontal by exerting a 43N force on the handle, which makes an angle alpha with the horizontal. The force of friction on the suitcase is 27N.

A) What angle does the handle make with the horizontal
B) What is the normal force on the suitcase
C) What is the coefficient of friction µs ?
friction = coefficient multiplied by the normal or f = µs N

 

11) extra for Physics major:

You push a 325N truck up a 20 degrees inclined plane at a constant velocity by exerting a 211N force parallel to the plane’s surface.

A) what is the component of the trunk’s weight parallel to the plane ?

B) What is the sum of all forces parallel to the plane’s surface ?

C) What are the magnitude and direction of the friction force ?
D) What is the coefficient of friction ? (f = µs N)