describe an experiment to verify the Law of parelellogram of force
Answers
Explanation:
The law of parallelogram of forces states that if two vectors acting on a particle at the same time be represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point their resultant vector is represented in magnitude and direction by the diagonal of the parallelogram drawn from the same point .
Parallelogram law of forces
law of parallelogram of forces
Okay, got it what it wants to say. But, the main question here is how does it happen! Right?
So let us assume that the two vectors A and B, inclined at angle θ, be acting on a particle at the same time. Let they be represented in magnitude and direction by two adjacent sides OP and OS of parallelogram OPQS, drawn from a point O.
According to parallelogram law of vectors , their resultant vector will be represented by the diagonal of the parallelogram .
Magnitude and Direction of Resultant:
Parallelogram law of forces
law of parallelogram of forces
Draw a perpendicular QN to OP produced.
And let us assume that OP=A, OS= PQ= B, OQ=R and angle SOP= angle QPN = θ.
Now considering this if we proceed further , in the case of triangle law of vector addition , the magnitude and direction of resultant vector will be given by
R= sqrt of A^2 + B^2 + 2 AB cosθ
tan B = B sinθ/ A+B cosθ
HOPE IT MAY HELP YOU
Answer:
The law of parallelogram of forces states that if two vectors acting on a particle at the same time be represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point their resultant vector is represented in magnitude and direction by the diagonal of the parallelogram drawn from the same point.
Explanation:
So let us assume that the two vectors A and B, inclined at angle θ, be acting on a particle at the same time. Let they be represented in magnitude and direction by two adjacent sides OP and OS of parallelogram OPQS, drawn from a point O.
According to the parallelogram law of vectors, their resultant vector will be represented by the diagonal of the parallelogram.
Draw a perpendicular QN to OP produced.
And let us assume that OP=A, OS= PQ= B, OQ=R, and angle SOP= angle QPN = θ.
Now considering this if we proceed further, in the case of the triangle law of vector addition, the magnitude and direction of the resultant vector will be given by
R= sqrt of A^2 + B^2 + 2 AB cosθ
tan B = B sinθ/ A+B cosθ