A 1190-N solid beam is 1.8 m long. It is attached to a wall at one end and is su

Question

A 1190-N solid beam is 1.8 m long. It is attached to a wall at one end and is supported by a cable attached to the beam at a distance of d0.27 m Given: -35 b21 34 o from the other end. A 1860-N crate hangs from the far end of the beam Find (a) the magnitude of the tension in the wire (b) the magnitudes of the horizontal and vertical components of the force that the wall exerts on the left end of the beam horizontal vertical the magnitude of the torque exerted by (just) the cable about the point where the beam connects to the wall N m

Explanation / Answer

forces acting on the beam are:

tension in the cable =T

weight of the beam ,acting at the middle of the beam

weight of the crate at the right end of the beam

force from the wall on the beam:

in horizontal direction=F1

in vertically upward direction=F2

balancing horizontal forces:

F1=T*cos(c)

==>F1=0.829*T…(1)

balancing forces along vertical direction:

F2+T*sin(c)=1190+1860

==>F2+0.5592*T=3050…(2)

balancing torque about the point where the beam meets the wall:

T*cos(b)*(1.8-0.27)=1190*(1.8/2)*cos(a)+1860*1.8*cos(a)

==>T=2534.2 N

using value of T,

we get

F1=2260.5 N

from equation 2,

F2=1632.9 N

so answers are:

part a:

magnitude of tension =2534.2 N

part b:

horizontal force=F1=2260.5 N

vertical force=F2=1632.9 N

part c:

torque exerted=T*cos(b)*(1.8-0.27)

=3619.8 N.m

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