A symmetrical table of height 0.730 m, length 1.75 m, and weight 433 N is dragge
ID: 1477712 • Letter: A
Question
A symmetrical table of height 0.730 m, length 1.75 m, and weight 433 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 0.425.
(b) Calculate the normal force and the frictional force on each leg.
Show that the table will not tip. (Do this on paper. Your instructor may ask you to turn in this work.)
Explanation / Answer
let minimum force necessary be F N.
then its vertical component=F*sin(30)=0.5*F, acting in vertically upward direction
normal force from ground=weight of the table-vertical component of dragging force
=433-0.5*F
friction force=friction coefficient*normal force
=0.425*(433-0.5*F)
=184.025-0.2125*F
hence the horizontal component of the dragging force should be atleast greater than friction force in order for the table to move
hence F*cos(30)=184.025-0.2125*F
==>0.866*F=184.025-0.2125*F
==>1.0785*F=184.025
==>F=170.63 N
part b:
assuming weight is symmetrically distributed on both the legs.
weight on each leg=433/2=216.5 N
vertical force on front leg=F*sin(30)=85.315 N
hence normal force on front leg=weight-vertical force on front leg=131.185 N
friction force=friction coefficient*normal force=0.425*131.185=55.753 N
weight on back leg=216.5 N
hence normal force on back leg=216.5 N
frictional force=friction coefficient*normal force
=0.425*216.5=92.0125 N
torque about front leg=horizontal component of drag force*height-weight on back leg *length
=170.63*cos(30)*0.73-216.5*1.75
=-217 N.m
hence it is in anticlockwise direction
and the table will not tip by the drag force.
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