Two forces of magnitude 10N & 8N are acting at a point. If the angle b/w the two forces is 60°, how do I determine the magnitude of the resultant force?
Since typing all those formulas were getting cumbersome I just solved it on a piece of paper and uploaded it.For reference,I have taken vector A=10N, vector B=8N, the angle between vectors A & B asθ (theta) = 60 degrees.The resultant vector is C and at an angle ø (phi) with vector A.Notice another thing. I have taken ø between vector A and C. If I need to know the angle between vector B and C, say angle α (alpha)or just α = 60 degrees - θHope that helps.
Two or more forces acting on the same point at the same time are called?
1 . Two or more forces acting on the same point at the same time are called A. concurrent forces. B. applied forces. C. normal forces. D. parallel forces. 2 . A force acting perpendicular to the surface is called the A. concurrent force. B. applied force. C. normal force. D. frictional force. 3 . The force that always opposes motion is the A. normal force. B. applied force. C. parallel force. D. frictional force. 4 . The sum of all forces acting in the same direction is called the A. resultant force. B. applied force. C. net force. D. parallel force.
Two concurrent forces of 40 N and X N have a resultant of 100N. Force X could be?
(C) 80. The forces have to be acting at an angle to one another (from 0 to 180 degrees, let's say). (A) 20 can't be right because even combined at the same direction (0 degree angle), 40 + 20 is only sixty. Same with (B) 40. 40 + 40 < 100. And at other extreme, even if the forces were 180 degrees apart (the maximum) 150 - 40 is 110 and too high, so it can't be (D). That leaves only (C).
What is the resultant force when the two forces act at angles to each other.Degrees=60,One force acting with force of 3N and one with a force of 4N?
We need to use the Law of parallelogram of vectors.The law states that when two vectors (forces) act on a particle at the same time are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point their resultant vector is represented by the diagonal of the parallelogram in magnitude and direction drawn from the same point.i m telling u the result if two forces A and B are acting at an angle of 60 degree then the magnitide of their resultant will beJust plug in a = 3 ,b = 4 and x =60degrees to get ur answer as sqrt(37)
Physics: What is the magnitude and direction of a third force?
this question can be solved by resolution of vectors we need to find the resultant of the two forces and than the third force will be opposite in direction and equal in magnitude to the resultant. x components F1x=0(at 90') F2x=44cos60=22 y components F1y=33sin90= 33N F2y=44sin60=38.10 Rx=22N Ry=33+38.1=71 R^2=(Rx)^2+(Ry)^2 R=72N angle will be tan(inverse)Ry/Rx Q=72.7 the angle should be in the 3rd quardrant and will be 252
What is the vector sum of a 65N force acting due east and a 32N force acting due west?
65N + (-32N) = 33N due east
Why does the apparent weight of a body decrease in a lift moving downwards?
The apparent weight of the body decreases only when the lift is accelerating or else their is no change.The reason for feeling weight:When we exert a force on the floor of the lift i.e. our weight, according to Newton's third law the floor of the lift exerts an equal and opposite force on us. This reaction force is felt as the apparent weight and is called normal reaction force.You need to be familiar with the term 'Pseudo Force'.Pseudo Force: Any force that is postulated to account for apparent deviations from Newton's laws of motion appearing in an accelerated reference system.Pseudo force always acts opposite to the direction of acceleration.Let's say the lift is moving downwards with an acceleration 'a'.let's consider the forces acting on the floor of the lift.A force due to weight of our body acts downwards=mg.Pseudo force acts upwards=ma (since the lift is accelerating downwards)Net force=mg-maTherefore, their is a net decrease in downward force and hence a net decrease in normal reaction force. So, the apparent weight of the body decreases.Note: The above problem has been solved with reference to the frame of the lift, which is a non-inertial frame.