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**Problem 3.156 10 lb A couple of magnitude 30 lb**

12 in 8 in 60o 45 lb 30 lb 10 lb A couple of magnitude M = 54 lb.in. and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resultant intersects line AB and line BC.

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**Solving Problems on Your Own**

A couple of magnitude M = 54 lb.in. and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resultant intersects line AB and line BC. Problem M A B C 12 in 8 in 60o 45 lb 30 lb 10 lb 1. Determine the resultant of two or more forces. Determine the rectangular components of each force. Adding these components will yield the components of the resultant.

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**M, obtained by adding the moments about the point of the various**

Solving Problems on Your Own A couple of magnitude M = 54 lb.in. and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resultant intersects line AB and line BC. Problem M A B C 12 in 8 in 60o 45 lb 30 lb 10 lb 2. Reduce a force system to a force and a couple at a given point. The force is the resultant R of the system obtained by adding the various forces. The couple is the moment resultant of the system M, obtained by adding the moments about the point of the various forces. R = S F M = S ( r x F )

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**Solving Problems on Your Own**

A couple of magnitude M = 54 lb.in. and the three forces shown are applied to an angle bracket. (a) Find the resultant of this system of forces. (b) Locate the points where the line of action of the resultant intersects line AB and line BC. Problem M A B C 12 in 8 in 60o 45 lb 30 lb 10 lb 3. Reduce a force and a couple at a given point to a single force. The single force is obtained by moving the force until its moment about the point (A) is equal to the couple vector MA. A position vector r from the point, to any point on the line of action of the single force R must satisfy the equation r x R = MA R

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**S F : R = ( _ 10 j ) + (_ 45 i ) + ( 30 cos 60o i + 30 sin 60o j ) **

M A B C 12 in 8 in 60o 45 lb 30 lb 10 lb Problem Solution Determine the resultant of two or more forces. (a) Adding the components of the forces: S F : R = ( _ 10 j ) + (_ 45 i ) + ( 30 cos 60o i + 30 sin 60o j ) = _ ( 30 lb ) i + ( lb ) j R = 34.0 lb o

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**S MB : MB = ( 54 lb.in ) + ( 12 in )(10 lb) _ ( 8 in ) ( 45 lb) **

A B C 12 in 8 in 60o 45 lb 30 lb 10 lb Problem Solution Reduce a force system to a force and a couple at a given point. (b) First reduce the given forces and couple to an equivalent force-couple system (R, MB) at B. + S MB : MB = ( 54 lb.in ) + ( 12 in )(10 lb) _ ( 8 in ) ( 45 lb) = _ 186 lb.in

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**S MB : _ 186 lb.in = _ a ( 15.9808 lb) or a = 11.64 in and with R at E **

C a c R D E Problem Solution Reduce a force and a couple at a given point to a single force. With R at D: S MB : _ 186 lb.in = _ a ( lb) or a = in and with R at E S MB : _ 186 lb.in = c ( 30 lb) or c = 6.20 in + + The line of action of R intersects line AB in. to the left of B and intersects line BC 6.20 in below B.

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