Up to this point, we have only looked at forces and moments in 2 ?dimensions.? Adding the 3rd dimension complicates our calculations.? We ?have to break the 3D problem into two 2D pro
Up to this point, we have only looked at forces and moments in 2 dimensions. Adding the 3rd dimension complicates our calculations. We have to break the 3D problem into two 2D problems and analyze each separately.
Friction is a special case of static problems where the friction force depends on the types of surfaces in contact where the sliding will occur and the weight of the object to which a force is applied.
complete and submit the problems below
MET211 Statics
Friction
Friction Laws for Dry Surfaces
The following discussion will concern only nonlubricated surfaces.
Motion or impending motion of two surfaces in contact causes a reaction force known as a friction force, F.
This friction force is:
1. Parallel to a flat surface or tangent to a curved surface
2. Opposite in direction to the motion or impending motion
3. Dependent on the force pressing the surfaces together
4. Generally independent of the area of surface of contact
5. Independent of velocity, except for extreme cases not to be considered here
6. Dependent on the nature of the contacting surfaces
Friction Laws for Dry Surfaces
Impending motion means that the object being considered is on the verge of moving
A small additional force would cause motion.
An object said to be in impending motion is not moving, and therefore is in static equilibrium.
A free-body diagram showing all external forces can be drawn.
Coefficients of Friction
Consider the case of a block on a horizontal surface being pushed by a force P so that the block has impending motion to the right.
P: as shown W: weight of the block F: force of friction N: normal force
Coefficients of Friction
When the friction force reaches its maximum value, we still have static equilibrium but the block is on the verge of moving and therefore has impending motion.
In order to describe the friction between two surfaces, we use the relationship between the friction force and the normal force.
The friction force depends on the normal force and is always a fraction of the normal force.
The friction force is therefore expressed as a fraction or portion of the normal force.
This relationship is called the coefficient of friction and is represented by the Greek lowercase letter mu (μ).
Angle of Friction
A second way in which friction is described is by the angle of friction.
In the figure to the right, the friction force Fmax and the normal force N are combined in the resultant R.
The friction angle ϕ (phi) is the angle between N and R.
Considering the right-angle triangle in which ϕ is located, we can write
So
Remember
Angle of Friction
Example:
The 80-N force shown in the figure causes impending motion to the right.
Determine the static coefficient of friction μ
Solution:
N = 400 N↑ and
Belt Friction
The two main assumptions in this section are:
1. A rope, cable, or flat belt is used. (Using a notched belt or a V-belt would require further analysis.)
2. Motion is impending—in the manner of a rope wound around a fixed cylinder so that the rope is starting to slip.
Belt Friction
Consider the simplified case of a rope passed over a fixed cylindrical beam and a force P causing impending motion of the weight upward.
Due to friction, force P will be larger than weight W.
The rope has two different tensions, a large tension (TL) and a small tension (TS).
Belt Friction
The free-body diagram of the portion of rope passing over the cylinder is shown in figure below.
Notice that the friction force is acting in a direction opposite to that of the impending motion of the rope with respect to the cylinder.
A free-body diagram of the rope shows the force of friction on the rope due to the cylinder.
There is impending motion of the cylinder with respect to the rope, so we show the force of the cylinder on the rope.
Belt Friction
For impending slipping, the difference between TL and TS depends on:
1. Coefficient of friction μ
2. Angle of contact θ
The governing equation is
OR
Belt Friction
Example:
The 50-kg mass has impending motion upward when P = 750 N. Determine the coefficient of friction between the rope and the cylinder.
TL = P = 750 N
and
T s = 9.81 ( 50 ) = 490 N
θ = 90° = π/2 rad
Belt Friction
Example:
The 50-kg mass has impending motion upward when P = 750 N. Determine the coefficient of friction between the rope and the cylinder.
Solution Hints
1. The direction of the friction force is always opposite to the impending motion of the object of which you have drawn a free-body diagram. For flat surfaces it is parallel to the surface, and for curved surfaces it is normal to the radius at the point of contact.
2. The friction angle ϕ is between the resultant and the normal force.
3. If motion is not impending, then μ ≠ Fmax/N and F and N must be treated as any other two unknowns.
4. Don’t draw a vector triangle without first drawing a free-body diagram showing the same forces.
5. When solving for TL or TS in the equation ln(TL/TS) = μθ, keep in mind that (TL/TS) must be treated as a single term until both sides of the equation are antilogged.
That is, ln TL ≠ TS(μθ). Also remember that θ is in radians.
References
All figures and examples are taken from
APPLIED MECHANICS FOR ENGINEERING TECHNOLOGY, EIGHTH EDITION, Keith M. Walker, ISBN 978-0-13-172151-7
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MET211 Statics
Three Dimensional Equilibrium
RESULTANT OF PARALLEL FORCES
A system of parallel forces in three dimensions is shown below.
Gravity is the usual source of parallel forces.
Applications:
The design of column foundation pads
As in two-dimensional statics, a resultant is a single force that has the same effect as the system of forces that it replaces.
The magnitude of the resultant is equal to the algebraic sum of the vertical forces.
RESULTANT OF PARALLEL FORCES
In three-dimensional analysis, the axis appears to be a point. The forces shown in the figure have moments about both the x- and z-axes, but not about the y-axis.
Take moments about the x-axis and using the sign convention of
clockwise moments as negative and
counterclockwise moments as positive
RESULTANT OF PARALLEL FORCES
To visualize the moment direction, imagine yourself looking down the z-axis toward the origin.
RESULTANT OF PARALLEL FORCES
To visualize the moment direction, imagine yourself looking down the x-axis toward the origin.
RESULTANT OF PARALLEL FORCES
Example:
Determine the magnitude and location of the force system shown in Figure. The grid dimension is 1 m.
RESULTANT OF PARALLEL FORCES
Solution:
Step 1. Calculate the resultant, R
Step 2. Draw the side view or z-y plane
Step 3. Equate the moment of R (Rz̅) to all other moments using the moment sign convention.
Step 4. Draw the front view or x-y plane
Step 5. Equate to the other moments.
RESULTANT OF PARALLEL FORCES
Solution:
Step 1. Calculate the resultant, R
RESULTANT OF PARALLEL FORCES
Solution:
Step 2. Draw the side view or z-y plane
RESULTANT OF PARALLEL FORCES
Solution:
Step 3. Equate the moment of R to all other moments using the moment sign convention.
2.9
RESULTANT OF PARALLEL FORCES
Solution:
Step 4. Draw the front view or x-y plane
RESULTANT OF PARALLEL FORCES
Solution:
Step 5. Equate to the other moments.
Rx̅
2.9 m
EQUILIBRIUM OF PARALLEL FORCES
The direction and location of forces are known, so it becomes a matter of writing moment equations about the correct points or axes.
The moment equations give simultaneous equations with two unknowns.
EQUILIBRIUM OF PARALLEL FORCES
Example:
A horizontal plate is represented by the grid in Figure below, where each square has sides 1 ft in length. The plate is supported at A, B, and C and has 20 lb applied as shown. Neglect the weight of the plate and determine the reactions at A, B, and C.
EQUILIBRIUM OF PARALLEL FORCES
Solution:
Draw a side view by looking down the x-axis toward the z-y plane. This is often referred to as “projecting the forces into the z-y plane.”
Draw a front view by looking down the z-axis and viewing the forces in the x-y plane.
Choose one of the three unknown forces—A, for example—and take moments about the point through which it passes in each diagram.
Simultaneous equations with the unknowns B and C are then solved.
EQUILIBRIUM OF PARALLEL FORCES
Solution:
Draw a side view by looking down the x-axis toward the z-y plane. This is often referred to as “projecting the forces into the z-y plane.”
EQUILIBRIUM OF PARALLEL FORCES
Solution:
Draw a front view by looking down the z-axis and viewing the forces in the x-y plane.
EQUILIBRIUM OF PARALLEL FORCES
Solution:
Simultaneous equations with the unknowns B and C are then solved.
For the side view :
∑MA = 0
( 3 ft ) B − ( 20 lb ) ( 1 ft ) − ( 2 ft ) C = 0
3 B − 2 C = 20
For the front view:
∑MA = 0
( 2 ft ) B + ( 4 ft ) C − ( 20 lb ) ( 1 ft ) = 0
2 B + 4 C = 20
EQUILIBRIUM OF PARALLEL FORCES
COMPONENTS AND RESULTANTS OF FORCES IN SPACE
The sign convention for each of the x-, y-, and z-axes is shown in Figure
COMPONENTS AND RESULTANTS OF FORCES IN SPACE
Rather than adding the components in two steps, we will solve for R in one step
COMPONENTS AND RESULTANTS OF FORCES IN SPACE
Example:
COMPONENTS AND RESULTANTS OF FORCES IN SPACE
Example:
Resolve a force in space into components in the x-, y-, and z-directions.
Suppose that a 100-lb force in space has coordinates of (8, 4, 2)
COMPONENTS AND RESULTANTS OF FORCES IN SPACE
Example:
Resolve a force in space into components in the x-, y-, and z-directions.
Suppose that a 100-lb force in space has coordinates of (8, 4, 2)
COMPONENTS AND RESULTANTS OF FORCES IN SPACE
Example:
Resolve a force in space into components in the x-, y-, and z-directions.
Suppose that a 100-lb force in space has coordinates of (8, 4, 2)
The x dimension of the box (8) represents
8/9.16 × 100 = 87.4 lb.
EQUILIBRIUM IN THREE DIMENSIONS
Forces in space are noncoplanar, and may be one of the following:
1. Parallel
2. Nonconcurrent: When the forces in three dimensions do not intersect at a common point, they form a noncoplanar, nonconcurrent force system
3. Concurrent : The compressive loads in the legs of a camera tripod
For forces in three dimensions, we have
∑ F x = 0 ∑ M x = 0
∑ F y = 0 ∑ M y = 0
∑ F z = 0 ∑ M z = 0
In a three-dimensional noncoplanar force system, moments are taken about an axis.
NONCONCURRENT, THREE-DIMENSIONAL SYSTEMS
Example:
The crank in the figure has a smooth bearing at B and a ball and socket at D. Calculate all reaction components at B and D.
Note:
A ball and socket can support forces in three directions, that is, Dx, Dy, and Dz, but no moments.
NONCONCURRENT, THREE-DIMENSIONAL SYSTEMS
NONCONCURRENT, THREE-DIMENSIONAL SYSTEMS
NONCONCURRENT, THREE-DIMENSIONAL SYSTEMS
NONCONCURRENT, THREE-DIMENSIONAL SYSTEMS
CONCURRENT, THREE-DIMENSIONAL SYSTEMS
Example:
Determine the tensile load in cable DB and the compressive loads in members AB and BC Neglect the weights of the members.
We will solve this problem by the view method of projecting the forces on a plane.
Find the length of diagonal DB.
CONCURRENT, THREE-DIMENSIONAL SYSTEMS
Example:
Determine the tensile load in cable DB and the compressive loads in members AB and BC Neglect the weights of the members.
We will solve this problem by the view method of projecting the forces on a plane.
Find the length of diagonal DB.
CONCURRENT, THREE-DIMENSIONAL SYSTEMS
CONCURRENT, THREE-DIMENSIONAL SYSTEMS
∑Mz = 0
CONCURRENT, THREE-DIMENSIONAL SYSTEMS
∑MA = 0
CONCURRENT, THREE-DIMENSIONAL SYSTEMS
∑MC = 0
References
All figures and examples are taken from
APPLIED MECHANICS FOR ENGINEERING TECHNOLOGY, EIGHTH EDITION, Keith M. Walker, ISBN 978-0-13-172151-7
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6-1. Determine the magnitude and location of the resultant of the force systems shown in Figures P6–1.
6-3. Determine the magnitude and location of the resultant of the force systems shown in Figures P6–3.
6–9. A platform supported by three ropes (shown in Figure P6–9) has a mass of 204 kg and carries a load of 4 kN. Calculate the tension in each rope.
6–10. A carport roof is supported as shown in Figure P6–10. If the roof has a mass of 200 kg, determine the load on each support.
6–13. Determine the resultant of the forces shown in Figure P6–13. Find the coordinates.
6–15. An anchor block is acted upon by the forces shown in Figure P6–15. Determine the resultant force.
7–1. A horizontal force of 80 N is required to start a 30-kg block sliding along a horizontal surface. What is the coefficient of static friction?
7–2. Determine the mass of a block if a force of 15 N is required to start it sliding on a horizontal surface (μ = 0.40).
7–12. Determine the minimum force P that will cause impending motion of the block shown in Figure P7–12. Will the block tip or slide? (If tipping occurs about point A, both the normal force N and the friction force F will be acting through point A.)
7–38. Determine the force T that will produce impending motion upward of the 500-lb weight in Figure P7–38. (Assume a coefficient of friction of 0.23.) Also find the minimum force T required to hold the 500-lb weight.
7–39. Determine the mass that can be lifted by a rope with three-quarters of a turn around a fixed horizontal shaft (μ = 0.3) if a force of 8 kN is applied to its other end.
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