A symmetrical table of height 0.780 m, length 1.60 m, and weight 432 N is dragge
ID: 1472620 • Letter: A
Question
A symmetrical table of height 0.780 m, length 1.60 m, and weight 432 N is dragged across the floor by a force applied to its front edge. The force is directed to the right and upward and makes an angle of 30.0° with the horizontal. (a) Find the minimum force necessary to drag the table across the floor. The coefficient of sliding friction between table and floor is 0.415. (b) Calculate the normal force and the frictional force on each leg.
front legs: normal force
frictional force
back legs: normal force
frictional force
Show that the table will not tip.
Explanation / Answer
along vertical
N + F*sintheta = W
N = W - F*sintheta
fx = u*N
along horizantal
Fx = F*costheta
Fx = fx
F*costheta = u* (W - F*sintheta)
F*cos30 = 0.415*(432-(F*sin30))
F = 167 N
------
In equilibrium
net torqe about the back legs = 0
F*cos30*h/2 + W*L/2 = F*sin30*L/2 + N1*L
(167*cos30*0.38) + (432*0.8) = (167*sin30*0.8)+(N1*1.6)
N1 = 208.6 N <<---answer
frictional force = f1 = u*N1 = 0.415*208.6 = 86.6 N <<---answer
along verticla Fnet = 0
N1 + N2 + F*sin30 = W
208.6 + N2 + 167*sin30 = 432
N2 = 139.9 N <<------answer
frictional force f2 = u*N2 = 0.415*139.9 = 58.06 N <<---answer
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