NAME & ID DATE MTE 119 STATICS HOMEWORK 3 SOLUTIONS
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1 NAME & ID DATE MTE STATICS HOMEWORK Problem : Textbook Exercise -4 Goal: Determine the magnitudes of the forces, and for equilibrium of the particle. Write the forces in Cartesian system and then use the equilibrium equations for the particle in x, y and z axis to find the magnitudes. Writing orces in Cartesian form: iˆ {cos 0 sin 0 kˆ} {0.5 iˆ 0.8 k ˆ = = } N iˆ 4 ˆ} j {0. iˆ 0.8 ˆ = { = j} N 5 5 iˆ = cos0 sin 0 ˆ} j = { 0.8 iˆ 0.5 ˆj } N { Equilibrium Equations for particle: or the x-axis: x or the y-axis: y = or the z-axis: z = sin cos0 Solving the above system of equations: = 800N, = 4N, = 54N 0
2 NAME & ID DATE MTE STATICS HOMEWORK Problem : Textbook Exercise -5 Cables A and can sustain a maximum tension of 500 N, and the pole can support a maximum compression of 00 N. Goal: Determine the maximum weight of the lamp that can be supported in the position shown. Assumptions: The force in the pole acts along the axis of the pole. irst write the forces and the weight of the lamp in Cartesian form, then write the equilibrium equations for point A and solve for the weight. Writing forces in Cartesian form:.5 AO = AO { iˆ ˆj k ˆ } N A = A { iˆ ˆj k ˆ } N = { iˆ ˆj k ˆ } N W = {Wk ˆ } N Equilibrium Equations: or the x-axis x AO A.5 or the y-axis.5 x AO A.5
3 NAME & ID DATE MTE STATICS HOMEWORK or the z- axis x AO A W.5 Assuming that one of the members is already under maximum load, we can solve the above equations to calculate what the load is in the other members. irst Case: A = 500N.5 AO AO (500) (500) AO (500) W = 88.N, AO = 444.4N > 00N Not acceptable Second case:.5 AO AO = 500N A AO A A (500) (500) (500) W A = 4.85N > 500N, AO = 85.4N > 00N Not acceptable Third case: (00).5.5 (00).5 (00).5 AO 0N A A A W = 80.8N, A 4N, W = 8N Answer
4 NAME & ID DATE MTE STATICS HOMEWORK 4 Problem : Textbook Exercise -5: Determine the tension developed in cables and and the strut, required to support the 50-kg crate. The spring OA has an un-stretched length of 0.8 m and a stiffness k OA =. kn / m. The force in the strut acts along the axis of the strut. irst draw the free body diagram and write the equilibrium equation for point O. Weight and spring force are known and the other can be found by solving equilibrium equations. ree ody Diagram:
5 NAME & ID DATE MTE STATICS HOMEWORK 5 Spring orce: sp = ks =.( 0.8).4kN = 40N Writing forces in Cartesian form: = = ( ( ˆ i 4 ˆj 4kˆ ( ) ( 4) 4ˆ i kˆ ( 4) 4 ) = 4 5 ˆ i 4 ˆj 4kˆ = ( ) = 4 4 sp = {40 ˆj } N ) = W = { 50(.8) kˆ} N = { 40.5k ˆ } N iˆ 5 iˆ Equilibrium equations: W sp iˆ kˆ ˆ j ˆ j kˆ kˆ ( 4 5 )ˆ i ( 4 ( ) 5 ( 40) ( 40.5) 5 ( Solving the above equations, the forces are: N, = 50N, = 480N 40) ˆj ) kˆ
6 NAME & ID DATE MTE STATICS HOMEWORK Problem 4: Textbook Exercise 4-4 and 4-5: 4-4 Take = 40 lb, C = 50lb and determine the moment of each force about the bolt located at A. 4-5 If lb and C = 45lb, determine the resultant moment about the bolt located at A. ind the force component normal to the rod and multiply it by it s distance from point O. or problem 4-4 Normal = cos 5 =.5lb M =.5(.5) = 0.lb. C Normal = C cos0 = 4.lb M = 4.(.5) = 4lb. or problem 4-5 M A cos 5 (.5) 45cos 0 (.5) = 5lb.
7 NAME & ID DATE MTE STATICS HOMEWORK Problem 5: Textbook Exercise 4-: The tool at A is used to hold a power lawnmower blade stationary while the nut is being loosened with the wrench at in the direction shown. Goals: Determine the moment it creates about the nut at C. What is the magnitude of force at A so that it creates the opposite moment about C? Use the fact that the summation of moments must be zero The magnitude of the force normal to the rod exerted by the wrench at : sin 0 = 50 sin 0 The magnitude of the moment of about C is: M C = sin 0 (0.) =. = N.m The magnitude of the moment exerted by the tool at A about C is: A MC = A( ) ( 0.4 ) N.m Since the two moments balance each other: A MC = MC :. = A( )(0.4) A = 5.N
8 NAME & ID DATE MTE STATICS HOMEWORK 8 Problem : Textbook Exercise 4-: The towline exerts a force of P=4kN at the end of the 0m long crane boom. If θ Goal: Determine the placement x of the hook at A so that this force creates a maximum moment about point O. What is this moment? The maximum moment occurs when the force is normal to the crane boom. ree-ody Diagram: 0 m.5 m D O 0 o 0 o x C A y geometry: L L DC C tan 0 x = L cos0 =.5 0sin 0 = L L L C DC =.m L =.5m =.m =.5 = 4m The maximum moment of P about O is then: M = P L = 4000(0) = 80kN.m O MAX
9 NAME & ID DATE MTE STATICS HOMEWORK Problem : Textbook Exercise 4-5: If a torque or moment of 80lb.in is required to looser the bolt at A, determine the force P that must be applied perpendicular to the handle of the flex-headed ratchet wrench. Since the force is perpendicular to the handle the resultant moment is the force magnitude multiplied by the distance from the bolt at A. M A = P(0.5 0sin 0 P = 80.4 = 8.5lb ) = 80
10 NAME & ID DATE MTE STATICS HOMEWORK 0 Problem 8: Textbook Exercise 4-4: The resultant couple moment created by the two couples acting on the disk is M = {0 kˆ }kip.in. R Goal: Determine the magnitude of force T. L Let us do: L = in. L = ( 4 ) in. in. L The resultant couple moment is: M R i= i M kip = TL TL = T() T() = T R = M The magnitude of the force T: T = 0.0kip
11 NAME & ID DATE MTE STATICS HOMEWORK Problem : Textbook Exercise 4-85: Two couples act on the frame. If d=4, determine the resultant couple moment. Compute the result by resolving each force into x and y components and a) inding the moment of each couple (Eq. 4-) b) Summing the moments of all the force components about point. R 4 R a) inding the moments of each couple: R = 4ˆ j 4 ˆ ˆ =80( ) i 80( ) j 5 5 R = ˆ i 50 sin 0 iˆ = 50 cos 0 jˆ C = i i i= M ( R ) M C i j k i j k = = { kˆ }lb. 50 sin 0 50 cos b) ind the component of the forces that creates a moment about and multiply by the distance from. 4 4 M C = 50 cos0 () 50cos0 (5) (80)() (80)(5) = lb. 5 5
12 NAME & ID DATE MTE STATICS HOMEWORK Problem 0: Textbook Exercise 4-: The gear reducer is subject to the couple moments shown. Determine the resultant couple moment and specify its magnitude and coordinate direction angles. Write the couples in Cartesian system and calculate the sum of them to obtain the resultant couple moment. ind the couple unit vector to find the coordinate direction angles. The moments in Cartesian form: M = {0ˆ} i lb. M iˆ ˆ = 80( cos0 sin 45 cos0 cos45 j sin0 kˆ) = { 48.ˆ i 48. ˆj 40kˆ} lb. Resultant Moment: M R = M M R = {.0ˆ i 4 ˆj 40kˆ} lb. Magnitude of the resultant moment: M R =.0 ( 48.) ( 40) = 4.lb. Coordinate direction angles: αβγ,, 0 α = cos β = cos γ = cos.0 ( ) = ( ) = ( ) = 4.
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