Unit 1 | unit 1 introduction to engineering mechanics

Unit - 1

Introduction to Engineering Mechanics

Q1) Find the magnitude of the two forces, such that if they act at right angles, their resultant is 10 N. But if they Act at 60°, their resultant is 13 N.

A1)

Given Two forces = and

First of all consider the two forces acting at right angles. We know that when the two given forces is 90 then the resultant force (R).

…(squaring both sides)

Similarly, when the angle between the two forces is 60, then the resultant force (R)

…(squaring both sides)

(Substituting )

We know that

….(i)

Similarly,

….(ii)

Solving equations (i) and (ii),

Q2) A triangle ABC has its side AB = 40 mm along positive x-axis and side BC = 30 mm along positive y-axis. Three forces of 40 N, 50 N and 30 N act along the sides AB, BC and CA respectively. Determine magnitude of the resultant of such a system of forces.

A2)

The system of given forces is shown in Fig 2.3.

From the geometry of the figure, we find that the triangle ABC is a right angled triangle, in which the side AC=50 mm. Therefore,

And

Resolving all the forces horizontally (i.e., along AB)

and now resolving all the forces vertically (i.e., along BC)

We know that magnitude of the resultant force,

Q3) A system of forces are acting at the corners of a rectangular block as shown in Fig. Determine the magnitude and direction of the resultant force.

A3)

Given System of forces

Magnitude of the resultant force

Resolving forces horizontally,

And now resolving the forces vertically

Magnitude of the resultant force

Q4) The following forces act at a point:

(i) 20 N inclined at 30° towards North of East,

(ii) 25 N towards North,

(iii) 30 N towards North West, and

(iv) 35 N inclined at 40° towards South of West.

Find the magnitude and direction of the resultant force.

A4) The system of the forces described can be drawn as:

Magnitude of the resultant force

Resolving all the forces horizontally i.e., along East-West line,

and now resolving all the forces vertically, i.e., along North-South line,

We know the magnitude of the resultant force,

Direction of the resultant force

Let θ= Angle, which the resultant force makes with the East.

Since is negative and is positive, therefore resultant lies between 90 and 180. Thus actual angle of the resultant =180-47.7=132.3

Q5) A horizontal line PQRS is 12 m long, where PQ = QR = RS = 4 m. Forces of 1000 N, 1500 N, 1000 N and 500 N act at P, Q, R and S respectively with downward direction. The lines of action of these forces make angles of 90°, 60°, 45° and 30° respectively with PS. Find the magnitude, direction and position of the resultant force.

A5) Magnitude of the resultant force

Resolving all the forces horizontally,

and now resolving all the forces vertically,

We know that magnitude of the resultant force,

Direction of the resultant force

Let θ=Angle, which the resultant force makes with PS

Note:

Since both the values of and are +ve. Therefore resultant lies between 0 and 90. Positive of the resultant force

Let Distance between P and the line of action of the resultant force.

Now, taking moments of the vertical components of the forces and the resultant force about P, and equating the same,

Q6) A uniform plank ABC of weight 30 N and 2 m long is supported at one end A and at a point B 1.4 m from A as shown in Fig. below. Find the maximum weight W, that can be placed at C, so that the plank does not topple.

A6) Weight of the plant ABC =30 N; Length of the plank ABC =2m and distance between end A and a point B on the plank (AB)=1.4m .

We know that weight of the plank (30 N) will act at its midpoint, as it is of uniform section.

This point is at a distance of 1 m from A or 0.4 m from B as shown in figure.

We also know that if the plant is not to topple, then the reaction at A should be zero for the maximum weight at C. now taking moments about B and equating the same.

Q7) Three forces of 2P, 3P and 4P act along the three sides of an equilateral triangle of side 100 mm taken in order. Find the magnitude and position of the resultant force.

A7)

Magnitude of the resultant of force

Resolving all the force horizontally,

And now resolving all the forces vertically.

We know that magnitude of the resultant force

Position of the resultant force

Let Perpendicular distance between B and the line of section of the resultant force.

Now, taking moments of the resultant force about B and equating the same,

Note: The moment of the force 2P and 4P about the point B will be zero, as they pass through it.

Q8) A square ABCD has forces acting along its sides as shown in Fig. below. Find the values of P and Q, if the system reduces to a couple. Also find magnitude of the couple, if the side of the square is 1 m.

A8)

Given Length of square = 1 m

Values of P and Q

We know that if the system reduce to a couple, the resultant force in horizontal and vertical directions must be zero. Resolving the forces horizontally,

100-100 cos 45 - P = 0

P = 100-100 cos 45 N

Now resolving the forces vertically,

Magnitude of the couple

We know that moment of the couple is equal to the algebraic sum of the moments about any point. Therefore moment of the couple (taking moments about A)

Since the value of moment is negative, therefore the couple is anticlockwise.

Q9) Determine the moment produced by the force F in Fig. shown about point O. Express the result as a Cartesian vector.

A9) As shown in Fig., either rA or rB can be used to determine the moment about point O. These position vectors are

and

Force F expressed as a Cartesian vector is

Thus

Q10) Two forces act on the rod shown in Fig. below. Determine the resultant moment they create about the flange at O. Express the result as a Cartesian vector.

A10)

Position vectors are directed from point O to each force as shown in fig 4-15b these vectors are

The resultant moment about O is therefore

Q11) Determine the required length of cord AC in Fig. below so that the 8-kg lamp can be suspended in the position shown. The undeformed length of spring AB is l = 0.4 m and the spring has a stiffness of kAB = 300 N/m.

A11)

By simply drawing the free-body diagram, we get,

In the force in spring AB is known, the stretch of the spring can be found using . From the problem geometry. It is then possible to calculate the required length of AC.

Free – Body Diagram

The lamp has a weight W=8(9.81) =78.5 N and so the free body diagram of the ring at a is shown in Fig3-8b.

Equation of Equilibrium

Using the x, y axes

Solving, we obtain

The stretch of spring AB is therefore

;

So the stretched length is

The horizontal distance from C to B fig 3-8a requires

Q12) Determine the force in each cable used to support the 40-lb crate shown in Fig.

A12)

Initially, free-body diagram is drawn below.

Equations of equilibrium:

First we will express each force in Cartesian vector form. Since the coordinates of points B and C are and We have

Equilibrium requires

Equating the respective i, j, k components to zero,

(1)

(2)

(3)

Equation (2) states that . Thus, solving Eq.(3) for and substituting the result into Eq.(1) to obtain . We have

Q13) Determine the tension in each cord used to support the 100-kg crate shown in Fig.

A13)

Free-Body diagram.

The force in each of the cords can be determined by investigating the equilibrium of point A. The free-body diagram is shown in Fig. 3-13b. The weight of the crate is

Equation of Equilibrium.

Each of force on the free-body diagram is first expressed in Cartesian vector form. Using Eq 2-9 for and noting point for . We have

Equilibrium requires

Equating the respective i, j, k components to zero,

(1)

(2)

(3)

Solving Eq.(2) for in terms of and substituting this into Eq. (3) yields . is then determined from Eq.(2). Finally, substituting the results into Eq.(1) gives . Hence,

Q14) A fixed crane has a mass of 1000 kg and is used to lift a 2400-kg crate. It is held in place by a pin at A and a rocker at B. The centre of gravity of the crane is located at G. Determine the components of the reactions at A and B.

A14)

Strategy:

Draw a free-body diagram to show all of the forces acting on the crane, then use the equilibrium equations to calculate the values of the unknown forces.

Modelling:

Free-body Diagram: By multiplying the masses of the crane and of the crate by , you obtain the corresponding weights-that is , 9810 N or 9.81N, and 23500 N or 23.5 kN (fig 1). The reaction at pin A is a force of unknown direction; you can represent it by components and . The reaction at the rocker B is perpendicular to the rocker surface; thus it is horizontal. Assume that and B act in the directions shown.

Analysis:

Determination of B. The sum of the moments of all external forces about point A is zero. The equation for this sum contains neither nor . Since the moments of and about A are zero. Multiplying the magnitude of each force by its perpendicular distance from A, you have

Since the result is positive, the reaction is directed as assumed.

Determination of :

Determine the magnitude of by setting the sum of the horizontal components of all external forces to zero.

Since the result is negative, the sense of is opposite to that assumed originally.

Determination of .

The sum of the vertical components must also equal zero. Therefore,

Adding the components and vectorially, you can find that the reaction at A is .

Q15) Three loads are applied to a beam as shown. The beam is supported by a roller at A and by a pin at B. Neglecting the weight of the beam, determine the reactions at A and B when P = 15 kips.

A15)

Strategy:

Draw a free-body diagram of the beam, then write the equilibrium equations, first summing forces in the x direction and then summing moments at A and at B.

Modelling:

Free-Body diagram:

The reaction at A is vertical and is denoted by A (fig.1 ). Represent the reaction at B by components and . Assume that each component acts in the direction shown.

Analysis:

Equilibrium Equations:

Write the three equilibrium equations and solve for the reactions indicated;

Q16) A loading car is at rest on a track forming an angle of 25° with the vertical. The gross weight of the car and its load is 5500 lb, and it acts at a point 30 in. from the track, halfway between the two axles. The car is held by a cable attached 24 in. from the track. Determine the tension in the cable and the reaction at each pair of wheels.

A16)

Strategy: Draw a free-body diagram of the car to determine the unknown force, and write equilibrium equations to find their values, summing moments at A and B and then summing forces.

Modelling:

Free-body diagram: The reaction at each wheel is perpendicular to the track, and the tension force T is parallel to the track. Therefore, for convenience, choose the x axis parallel to the track and the y axis perpendicular to the track (Fig. 1). Then resolve the 5500-lb weight into x and y components.

Analysis

Equilibrium Equations:

Take moments about A to eliminate T and from the computation.

;

Then take moments about B to eliminate T and , from the computation.

;

Determine the value of T by summing forces in the x direction.

Q17) The frame shown supports part of the roof of a small building. Knowing that the tension in the cable is 150 kN, determine the reaction at the fixed end E.

A17)

Strategy:

Draw a free-body diagram of the frame and of the cable BDF. The support at E is fixed, so the reactions here include a moment; to determine its value, sum moments about point E.

Modelling:

Free-body diagram:

Represent the reaction at the fixed end E by the force components and and the couple (fig 1.). The other forces acting on the free body are the four 20 kN loads and the 150 kN force exerted at end F of the cable.

Analysis:

Equilibrium Equations:

First note that

Then you can write the three equilibrium equations and solve for the reactions at E.

Q18) A 400-lb weight is attached at A to the lever shown. The constant of the spring BC is k = 250 lb/in., and the spring is unstretched when θ = 0. Determine the position of equilibrium.

A18)

Strategy:

Draw a free-body diagram of the lever and cylinder to show all forces acting on the body (Fig. 1), then sum moments about O. Your final answer should be the angle θ.

Modelling:

Free-Body Diagram: Denote by s the deflection of the spring from its unstretched position and note that . Then

Analysis:

Equilibrium Equation: Sum the moments of W and F about O to eliminate the reactions supporting the cylinder. The result is

Solving by trial and error, the angle is

Q19) Two blocks A and B of weights 1 kN and 2 kN respectively are in equilibrium position as shown in Fig. If the coefficient of friction between the two blocks as well as the block B and the floor is 0.3, find the force (P) required to move the block B.

A19) Given: Weight of block A (WA) = 1 kN; Weight of block B (WB) = 2 kN and

coefficient of friction (μ) = 0.3.

Forces acting on blocks A and B are plotted respectively as,

The forces acting on the two blocks A and B are shown in fig. 8.5(a) and (b) respectively .

First of all, consider the forms acting in block A

Resolve the forces vertically

And now resolving the forces horizontally

Diving equation 1 by 2

Now considered the block B. A little consideration will show that the downward force of the block A (equal to R) will also act along with the weight of the block B

Resolving the forces vertically

And now resolving the forces horizontally

Q20) What is the maximum load (W) which a force P equal to 5 kN will hold up, if the coefficient of friction at C is 0.2 in the arrangement shown in Fig. Neglect other friction and weight of the member. If W = 3 kN and P = 4.5 kN, what are the normal and tangential forces transmitted at C?

A20) Given force (p)=5KN nad coeffecient of friction at

Maximus load W

Let,

R= normal reaction of the pulley on the beam at C

First of all, consider the equilibrium of the beam AB. Taking moments about the hinge A and equation the same

R*1=5*1.5=7.5

Now consider the equilibrium of the pulley. it is subjected to a normal reaction of 7.5 KN (as calculated above) tends to rotate it. A little consideration will show that the rotation of the pulley is prevented by the frictional force between the pulley and beam at C. we know that maximum force of friction at C.

Now taking moment about the center of the pulley and equating the same

Normal and tangential force transmitted at C

Now considered a weight W equal to3KN suspended from the pulley and a force P equal to 4.5KN applied at B

Let ,

Again considered equilibrium of the beam taking moment about the hinge A and equating the same.

We know that the tangential force at C will be the frictional force at C will be the frictional force between the pulley and beam . Again taking the moment about the center of the pulley and equation the same

Q21) A body of weight 500 N is lying on a rough plane inclined at an angle of 25° with the horizontal. It is supported by an effort (P) parallel to the plane as shown in Fig. Determine the minimum and maximum values of P, for which the equilibrium can exist, if the angle of friction is 20°.

A21) Given weight of the body (W)=500N Angle at which plane is inclined

Maximum value of P

We know that for the minimum value of P, the body is the point of sliding downward. we also know that the body is the point of sliding downward then the force.

Q22) An inclined plane as shown in Fig. below. is used to unload slowly a body weighing 400 N from a truck 1.2 m high into the ground. The coefficient of friction between the underside of the body and the plank is 0.3. State whether it is necessary to push the body down the plane or hold it back from sliding down. What minimum force is required parallel to the plane for this purpose?

A22) Given weight of body (W)=400N and coeffecient of friction

Whether it is necessary to push the body down the plane or hold it back from sliding down

We know that

=400*0.894=357.9N

Force of friction

Now resolving the 400N force along the plane

We know that the force along the plane (which is responsible for sliding the body) is more than the force of friction, therefore the body will slide down. or in other words, it is not necessary to push the body down the plan e, rather it is necessary to hold it back from sliding down

Minimum force required parallel to plane

We know that the minimum force required parallel to plane to hold the body back

P=178.5-107.3=71.2 N Ans

Q23) It is observed that when the bed of the dump truck is raised to an angle of the vending machines will begin to slide off the bed, Fig. shown. Determine the static coefficient of friction between a vending machine and the surface of the truck bed.

A23) The free body diagram of the component will be as,

Equation in equilibrium

Since slipping impends at is referred to as the angle of repose and by comparison , it is equal to the angle of static friction notice from the calculation that is independent of the weight of the vending machine and so knowing provides a convenient method for determining the coeffecient of static friction .

Q24) The uniform 10-kg ladder in Fig. rests against the smooth wall at B, and the end A rests on the rough horizontal plane for which the coefficient of static friction is 0.3. Determine the angle of inclination of the ladder and the normal reaction at B if the ladder is on the verge of slipping.

A24) Free-Body Diagram. As shown on the free-body diagram, Fig. below, the frictional force must act to the right since impending motion at A is to the left.

Equation of equilibrium and friction =

Since the ladder is on the verge of slipping then . By inspection can be obtained directly.

Using the result . now can be found

Finally, the angle can be determined by summing moments about point A

Q25) Beam AB is subjected to a uniform load of 200N/m and is supported at B by post BC, Fig. below. If the coefficients of static friction at B and C are 0.2 and 0.5, determine the force P needed to pull the post out from under the beam. Neglect the weight of the members and the thickness of the beam.

A25) Free body diagram: The free body diagram is drawn as shown below.

Equation and equilibrium and friction

Post slips at B and rotates about C. This required

Using this result and solving Eqs.1 through 3, we obtain

P= 320N

Since >, slipping at C occurs. Thus, the other case of movement must be investigated

(Post slips at C and rotates about B). here

Solving eqs. 1 through 4 yields.

Q26) The turnbuckle shown in Fig. below has a square thread with a mean radius of 5 mm and a lead of 2 mm. If the coefficient of static friction between the screw and the turnbuckle is 0.25, determine the moment M that must be applied to draw the end screws closer together.

A26) The moment can be obtained by applying Eq. 8-3 since friction at two screws must be overcome, this requires

Here

W=2000N

r=5mm

Substituting these values intoEq.1 and solving gives

6374.7Nmm=6. 37N.m

Q27) Determine the required force P in Fig. 11–7a, needed to maintain equilibrium of the scissor’s linkage when 600. The spring is unstretched when 300 . Neglect the mass of the links.

A27) Free body is drawn as follows:

Virtual displacment :- The position coordinates , measured from the fixed point A, are used to locate . These coordinates are parallel to the lines of action of their corresponding forces. Expressing in terms of the angle using trignometry,

Differentiating , we obtain the virtual displacment of point B and D

Virtual work equation- force P does positive work since it acts in the positive sense of its virtual displacment . The spring force does negativity work since it acts opposite to its positivity virtual displacement . thus , the virtual -work equation becomes

Since

When

Q28) If the box in Fig. below has a mass of 10 kg, determine the couple moment M needed to maintain equilibrium when. Neglect the mass of the members.

A28) Initially, free body diagram is drawn as

Virtual Displacements: - The position coordinate , measured from the fixed-point B, locates the weight, 10(9.81) N. Here,

Where b is a constant distance. Differentiating this equation, we obtain

Virtual-Work Equation: - The virtual-work equation becomes

Substituting Eq.1 into this equation

Since , then

Since it is required that then

Q29) The mechanism in Fig. below supports the 50-lb cylinder. Determine the angle for equilibrium if the spring has an unstretched length of 2 ft when angle is 0. Neglect the mass of the members.

A29) The free body diagram is as shown in fig. below.

Virtual Displacements. The position coordinates are established from the fixed-point A to locate at D and at E. The coordinate , also measured from a, specifies the position of the 50-lb force at B. The coordinates can be expressed in terms of using trigonometry.