Consider a rigid 2.30-m-long beam (see figure) that is supported by a fixed 1.15

Question

Consider a rigid 2.30-m-long beam (see figure) that is supported by a fixed 1.15-m-high post through its center and pivots on a frictionless bearing at its center atop the vertical 1.15-m-high post. One end of the beam is connected to the floor by a spring that has a force constant k = 1200 N/m. When the beam is horizontal, the spring is vertical and unstressed. If an object is hung from the opposite end of the beam, the beam settles into an equilibrium position where it makes an angle of 23° with the horizontal. What is the mass of the object?

from the opposite end of the beam, the beam kg ng nded free-body diagram of the forces acting on the metry when determining the extended length of the Save Progress Practice Another Version

Explanation / Answer

if an object applies a force to a spring, then the spring applies an equal and opposite force to the object. Hooke’s law gives the force a spring exerts on an object attached to it with the following equation:

F = –kx

the minus sign shows that this force is in the opposite direction of the force that’s stretching or compressing the spring. The variables of the equation are: F which represents force, k which is called the spring constant and measures how stiff and strong the spring is, and x is the distance the spring is stretched or compressed away from its equilibrium or rest position.

According to above concept

mg=-Kx

m*-9.8=-1200*1.15

m=1380/9.8=140.8kg

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