am working on a problem that asks me to find out the acceleration of an object on a ramp, and I have drawn the first image. From it we could clearly see that a = g * sin (theta). But that formula does not work for all a-> because, as shown in the second image when a is larger, it is not a right triangle, so we cannot take a = g * sin (theta) to find a in this example, am I right?
Displacement, Velocity and Acceleration
In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.
Linear Displacement
The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.
I am working on a problem that asks me to find out the acceleration of an object on a ramp, and I have drawn the first image. From it we could clearly see that a = g * sin (theta). But that formula does not work for all a-> because, as shown in the second image when a is larger, it is not a right triangle, so we cannot take a = g * sin (theta) to find a in this example, am I right?
The formula will be the same, however the magnitude of acceleration will change due to change in angle theta.
You are correct! You can not take exact the acceleration value as g. Sin(theta). But formula will be the same.
Acceleration= g.Sin(theta), which takes several values of angle theta with increase in side ( a).
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How could the formula be the same? I mean it is not a right triangle.
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