Force X has a magnitude of 1260 pounds, and Force Y has a magnitude of 1530 pounds. They act on a single point at an angle of 4
5 degrees to each other. Find the magnitude of the equilibrant and the angle between the equilibrant and the 1260-pound force.
1 answer:
Answer:
Fe= 2579.68 P
α= 24.8°
Explanation:
Look at the attached graphic
we take the forces acting on the x-y plane and applied at the origin of coordinates
FX = 1260 P , horizontal (-x)
FY = 1530 P , forming 45° with positive x axis
x-y components FY
FYx= - 1530*cos(45)° = - 1081.87 P
FYy= - 1530*sin(45)° = - 1081.87 P
Calculation of the components of net force (Fn)
Fnx= FX + FYx
Fnx= -1260 P -1081.87 P
Fnx= -2341.87 P
Fny=FYy
Fny= -1081.87 P
Calculation of the components of equilibrant force (Fe)
the x-y components of the equilibrant force are equal in magnitude but in the opposite direction to the net force components:
Fnx= -2341.87 P, then, Fex= +2341.87 P
Fny= -1081.87 P P, then, Fex= +1081.87 P
Magnitude of the equilibrant (Fe)
Fe= 2579.68 P
Calculation of the direction of equilibrant force (α)
α= 24.8°
Look at the attached graphic
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Details are missing in the question. Complete text of the problem:
"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.
"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.
(a) Through what distance does it move before its release? (m)
(b) What are the magnitude and direction of the force the pitcher exerts on the ball? (Enter your magnitude to at least one decimal place.)"
Solution
(a) The pitcher accelerates the baseball from rest to a final velocity of , so
, in a time interval of
. The acceleration of the ball in the horizontal direction (x-axis) is therefore
And the distance covered by the ball during this time interval, before it is released, is:
(b) For this part we need to consider also the weight of the ball, which is
From this, we find its mass:
Now we can calculate the magnitude of the force the pitcher exerts on the ball. On the x-axis, we have
We also know that the ball is moving straight horizontally. This means that the vertical component of the force exerted by the pitcher must counterbalance the weight of the ball (acting downward), in order to have a net force of zero along the y-axis, and so:
(upward)
So, the magnitude of the force is
To find the direction, we should find the angle of F with respect to the horizontal. This is given by
From which we find
Answer:
4 (please see the attached file)
Explanation:
While the angular speed (counterclockwise) remained constant, the angular acceleration was just zero.
So, the only force acting on the bug (parallel to the surface) was the centripetal force, producing a centripetal acceleration directed towards the center of the disk.
When the turntable started to spin faster and faster, this caused a change in the angular speed, represented by the appearance of an angular acceleration α.
This acceleration is related with the tangential acceleration, by this expression:
at = α*r
This acceleration, tangent to the disk (aiming in the same direction of the movement, which is counterclockwise, as showed in the pictures) adds vectorially with the centripetal force, giving a resultant like the one showed in the sketch Nº 4.
