Presentation on theme: "Translational Equilibrium Physics Montwood High School R. Casao."— Presentation transcript:
Translational Equilibrium Physics Montwood High School R. Casao
Concurrent forces are those forces that are applied to or act on the same point, as shown at the point where T1, T2, and T3 meet. When two or more forces act at the same point, the resultant force is the sum of the forces applied at that point. The resultant force is the single force that has the same effect as the two or more forces that act together. When forces act in the same direction or opposite directions, the total force can be found by adding the forces that act in one direction and subtracting the forces that act in the opposite direction.
Equilibrium is the state of a body in which there is no change in its motion. A body is in equilibrium when the net force acting on it is zero; there is no acceleration (a = 0 m/s 2 ), the body is either at rest or is moving at a constant velocity. The study of objects in equilibrium is called statics. When two or more forces are acting at the same point, the equilibrant force is the force that when applied at that same point produces equilibrium. The equilibrant force is equal in magnitude to that of the resultant force but acts in the opposite direction.
If an object is in equilibrium in one dimension, the forces in one direction must equal the forces in the opposite direction. ΣF + direction = ΣF - direction
If 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. –Conditions for equilibrium: In general, to solve equilibrium problems: –Draw a free-body diagram from the point at which the unknown forces act. –Find the x- and y-components of each force. –Substitute the components in the equations: –Solve for the unknowns. This may involve two simultaneous equations.
Two Conditions for Equilibrium An object is said to be in equilibrium if and only if there is no resultant force and no resultant torque (rotation). –First condition (translational equilibrium): ΣF x = 0; ΣF y = 0 –The linear speed is not changing with time. Car at rest Constant speed
–Second condition (rotational equilibrium) Σtorque = 0; the object does not rotate, or rotates at a constant number of turns per unit time. –Torque is the tendency of a force to rotate an object around an axis; torque measures how hard something is twisted. Wheel at rest Constant rotation
Translational Equilibrium Only If all forces act at the same point, then there is no torque to consider and one need only apply the first condition for equilibrium: ΣF x = 0; ΣF y = 0 Example: Find the tension in ropes A and B. 80 N A B N A B 60 0 Free-body Diagram: ByBy BxBx
Read problem; draw sketch; construct a free-body diagram, indicating components. Choose x-axis horizontal and choose right direction as positive (+). There is no motion. The components B x and B y can be found from right triangle trigonometry. y x R
B x = B cos 60 0 B y = B sin N A B N A B 60 0 Free-body Diagram: ByBy BxBx
ΣF x = 0; B x – A = 0; B x = A B·cos 60 = A; two unknowns proceed to ΣF y = 0 ΣF y = 0; B y – 80 N = 0; B y = 80 N; B·sin 60 = 80 N Determine A: A = B·cos 60 = N·cos 60 = N 80 N A B 60 0 Free-body Diagram: ByBy BxBx 80 N A B sin 60 0 B cos 60 o BxBx ByBy F x = 0 F y = 0
Problem Solving Strategy 1.Draw a sketch and label all information. 2.Draw a free-body diagram. 3.Find components of all forces (+ and -). 4.Apply First Condition for Equilibrium: ΣF x = 0; ΣF y = 0 5. Solve for unknown forces or angles.
Example: Find Tension in Ropes A and B A B 400 N A B 1. Draw free-body diagram. 2. Determine angles AyAy ByBy AxAx BxBx 3. Draw/label components. Next we will find components of each vector.
Example: Find the tension in ropes A and B. F x = B x - A x = 0 F y = B y + A y - W = 0 B x = A x B y + A y = W A B W 400 N AyAy ByBy AxAx BxBx 4. Apply 1 st Condition for Equilibrium: First Condition for Equilibrium: F x = 0 ; F y = 0
Example: Find the tension in ropes A and B. B x = A x B y + A y = W A B W 400 N AyAy ByBy AxAx BxBx Using Trigonometry, the first condition yields: B cos 60 0 = A cos 30 0 A sin B sin 60 0 = 400 N A x = A cos 30 0 ; A y = A sin 30 0 B x = B cos 60 0 B y = B sin 60 0 W x = 0; W y = -400 N
Example: Find the tension in A and B. A B W 400 N AyAy ByBy AxAx BxBx B = · A We will first solve the horizontal equation for B in terms of the unknown A: B cos 60 0 = A cos 30 0 A sin B sin 60 0 = 400 N We now solve for A and B: Two Equations and Two Unknowns.
Example: Find Tensions in A and B. A · sin B · sin 60 0 = 400 N B = · A A · sin (1.732 A) · sin 60 0 = 400 N · A · A = 400 N 2 · A = 400 N A = 200 N A B 400 N AyAy ByBy AxAx BxBx B = A Now apply Trig to: A y + B y = 400 N A sin B sin 60 0 = 400 N
Example: Find B with A = 200 N. Rope tensions are: A = 200 N B = 346 N B = · A A = 200 N B = · (200 N) B = 346 N A B W 400 N AyAy ByBy AxAx BxBx
You may need to find the tension or compression in part of a structure, such as in a beam or a cable. Tension is a stretching force produced by forces pulling outward on the ends of an object. Compression is a force produced by forces pushing inward on the ends of an object.