A 12kg uniform board is wedged into a corner & held by a spring at a 55 deg angle. Bond length= 50 cm.k?

Question

A 12kg uniform board is wedged into a corner & held by a spring at a 55 deg angle. Bond length= 50 cm. spring constant is 180 N/m and is parallel to the floor.
a) What is the torque due to the weight
b) find the amount at which the spring is stretched from its unrestrained length
c)magnitude of the force exerted on the borded by the corner at the wedge

Answer 1

(a) T= R x 0.5 W

R=.05 L cos (55) = .143m (half the board length)
T= R (0.5 mg)
T=0.14 (0.5 x 12 x 9.81)
T=8.4 mN

(b)
F=k x

x=F/k

Sum of moments at teh wall= 0 = moment of the board against the wall - moment due to string

R1 x 0.5 mg cos(55) - R2 F =0
R2= .5 cos(55)=0.27m
F=(R1 x 0.5 mg cos(55) )/R2
F=8.4 m N / 0.27m= 31N
then x= F/k= 31N/180 N/m=0.172m

(c) Ft= Fb + Fs

Fb=W sin(55) = mg sin(55)=
Fb= 12 x 9.81 sin(55)=16.1

Fs= F cos(55)= 31 cos(55)=
Fs=17.8

Ft=286.3 N