Tutorial for Wk 1 Fri and Wk2 Mon Lectures
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Transcript of Tutorial for Wk 1 Fri and Wk2 Mon Lectures
Loughborough University The Wolfson School 13MMA101 Statics Tutorial 1
A.Gregory C.Eng [email protected] TW212
Question 1
.
In this problem we can break down the two forces into their horizontal and vertical components.
The beam runner is attached to the beam via wheels. As such, assuming that the wheels are frictionless, what can we say about the horizontal forces if we wish this to remain a statics problem?
What does this say about the direction of the resultant of the two forces?
α = 25°, determine the magnitude of the force P
Stage 1. Draw a vector triangle showing the two forces and a vertical resultant.
Loughborough University The Wolfson School 13MMA101 Statics Tutorial 1
A.Gregory C.Eng [email protected] TW212
Stage 2. Produce an equation relating the horizontal force due to the 1600N force
and that produced by P.
Stage 3. Solve this equation for P.
Calculate the magnitude of the resultant of the two forces?
a) Using cartesion components b) Using the cos rule
Loughborough University The Wolfson School 13MMA101 Statics Tutorial 1
A.Gregory C.Eng [email protected] TW212
If the vector P was adjusted to produce a resultant of 2500N, what would be the new magnitude and direction of P?
Hint: Use your equations from stage 2 and stage 3a.
Loughborough University The Wolfson School 13MMA101 Statics Tutorial 1
A.Gregory C.Eng [email protected] TW212
Question 2.
The forces shown in the diagram are acting on a particle which means that they are...C...............................? Because they are ...C....................................they cannot generate a turning .........................? Using vector notation where x=i and y=j calculate the directional vector in each case and then the unit vectors. Example: In the case of the 102N force we have directional values of -480i and 900j. This gives us a length of sqrt(480^2 +900^2)= 1020 in the direction of the 102N force. We then divide the i and j components by 1020 to produce a unit vector for the resultant=1: -(480/1020)i , +(900/1020)j We now convert these positional vectors into force vectors by multiplying by 102N i.e. –(480*102/1020)i, +(900*102/1020)j Calculate the i and j components for the other two forces.
Loughborough University The Wolfson School 13MMA101 Statics Tutorial 1
A.Gregory C.Eng [email protected] TW212
Force ID
Positional i vector
Positional j vector
Positional resultant
Force i component
Force j component
102 -480 900 1020 -48 90
106
200
Add the i components and then the j components. Total ..........................i N Total ..........................j N Now calculate the resultant (sqrt(i^2 + j^2))
Loughborough University The Wolfson School 13MMA101 Statics Tutorial 1
A.Gregory C.Eng [email protected] TW212
Question 3.
A variety of pulley systems are shown above. What can we say about the tension in a rope as it passes over a frictionless pulley? It’s tension is .............................................. Its line of action follows that of the .................................. Draw a free body diagram (FBD) of the lower pulley wheel pin for each case and calculate the value of T if the crate weighs 6000N i.e. 6kN.
a) b)
Loughborough University The Wolfson School 13MMA101 Statics Tutorial 1
A.Gregory C.Eng [email protected] TW212
c) d) e) Question 4 (The hard one)
An 1800-N load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in the position shown by a second cable CAD, which passes over the pulley A and supports a load P. Determine (a) the tension in cable ACB, (b) the magnitude of load P.
Loughborough University The Wolfson School 13MMA101 Statics Tutorial 1
A.Gregory C.Eng [email protected] TW212
What can we say about the tension in the cable ACB? What can we say about the tension in the cable CAD? If this is a statics problem, what can we say about the resultant of the forces on the pulley pin in the horizontal direction? Draw an FBD of the pulley pin through which Q acts.
Note how we have shown the direction and sign of the direction that we are resolving the forces Fx and Fy. Note also that we have set each equation to zero because this is a statics problem. Now solve for P.