Understanding the law of forces in parallelogram. click on the link toWatch the VIDEO explanation: Watch Video
Parallelogram Law Of Forces
The statement goes like this if the forces, acting at a point, be represented in magnitude and direction by two adjacent sides of a parallelogram drawn through that point of application, Their resultant is completely by the diagonal of the parallelogram drawn through that point.
If two forces acting at a point be represented in the magnitude and direction, then the resultant is completely represented by the diagonal of the parallelogram drawn through that point
Lets understand the proof of this law
Let P and Q be the two given forces acting at O, represented in magnitude and the direction by sides OA and OB respectively of a parallelogram OACB. Let angle AOB is equal to alpha be the angle at which these two forces act.
Alpha may be acute or obtuse. Click on the button Case 1
Case 1 When alpha is acute
The resultant R is represented in magnitude and direction by the diagonal OC of a parallelogram OACB
AC being equal and parallel to OB, represents in magnitude and direction the force Q.
Angle CAD is equal to angle BOA i.e., equal to Alpha Since they are the corresponding Angles
OD is equal to OA plus AD
Thus, we get OD is equal to P plus Q cos alpha. Let this be equation 1
Also we have CD is equal to Q sin alpha. Let this be equation 2
Click on the button Case 2
When alpha is obtuse
angle CAD is equal to 180 degrees minus angle BOA is equal to 180 degrees minus alpha
Therefore, OD is equal to P plus Q cos alpha. Let this be equation 3
Also we have CD is equal to Q sin alpha. Let this be equation 4
Thus from case 1 and 2 we have
OD is equal to P plus Q cos alpha
CD is equal to Q sin alpha.
To find the magnitude of resultant R applying the Pythagoras theorem to right angled triangle ODC, we get OC square is equal to OD square plus CD square
This Implies R square is equal to P plus Q cos alpha whole square plus Q sin alpha whole square substituting for OD and CD.
Therefore R is equal to root of P square plus Q square plus 2PQ cos alpha.
This gives the magnitude of the resultant R.
To find the angle between R and P
Let theta be the angle which resultant R makes with OA in the direction of P.
Therefore Sin theta is equal to Q sin alpha b y R and cos theta is equal to P plus Q cos alpha by R.
The value of theta which satisfies these two equations gives the angle between R and P.
Parallelogram Law Of Forces
The statement goes like this if the forces, acting at a point, be represented in magnitude and direction by two adjacent sides of a parallelogram drawn through that point of application, Their resultant is completely by the diagonal of the parallelogram drawn through that point.
If two forces acting at a point be represented in the magnitude and direction, then the resultant is completely represented by the diagonal of the parallelogram drawn through that point
Lets understand the proof of this law
Let P and Q be the two given forces acting at O, represented in magnitude and the direction by sides OA and OB respectively of a parallelogram OACB. Let angle AOB is equal to alpha be the angle at which these two forces act.
Alpha may be acute or obtuse. Click on the button Case 1
Case 1 When alpha is acute
The resultant R is represented in magnitude and direction by the diagonal OC of a parallelogram OACB
AC being equal and parallel to OB, represents in magnitude and direction the force Q.
Angle CAD is equal to angle BOA i.e., equal to Alpha Since they are the corresponding Angles
OD is equal to OA plus AD
Thus, we get OD is equal to P plus Q cos alpha. Let this be equation 1
Also we have CD is equal to Q sin alpha. Let this be equation 2
Click on the button Case 2
When alpha is obtuse
angle CAD is equal to 180 degrees minus angle BOA is equal to 180 degrees minus alpha
Therefore, OD is equal to P plus Q cos alpha. Let this be equation 3
Also we have CD is equal to Q sin alpha. Let this be equation 4
Thus from case 1 and 2 we have
OD is equal to P plus Q cos alpha
CD is equal to Q sin alpha.
To find the magnitude of resultant R applying the Pythagoras theorem to right angled triangle ODC, we get OC square is equal to OD square plus CD square
This Implies R square is equal to P plus Q cos alpha whole square plus Q sin alpha whole square substituting for OD and CD.
Therefore R is equal to root of P square plus Q square plus 2PQ cos alpha.
This gives the magnitude of the resultant R.
To find the angle between R and P
Let theta be the angle which resultant R makes with OA in the direction of P.
Therefore Sin theta is equal to Q sin alpha b y R and cos theta is equal to P plus Q cos alpha by R.
The value of theta which satisfies these two equations gives the angle between R and P.
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