A symmetrical table of height 0.860 m, length 1.75 m, and weight 436 N is dragge

Question

A symmetrical table of height 0.860 m, length 1.75 m, and weight 436 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. (please show work)

(a) Find the minimum force necessary to drag the table across the floor. The coefficient of sliding friction between table and floor is 0.410.

(b) Calculate the normal force and the frictional force on each leg.

front legs: normal forc? frictional force ?

back legs: normal force? frictional force?

Show that the table will not tip.

Explanation / Answer

(a) Let F be the minimum force necessary to drag the table. Let W be the weight of the table and N be the total normal reaction on the table.

Then,

Fcos30o = sN

Fsin30o + N = W

=> Fsin30o + Fcos30o/s = W

=> F = Ws / (ssin30o + cos30o) = 436 * 0.41 / (0.41sin30o + cos30o) = 166.9 N

Also, N = (Fcos30o) / s = 166.9 * cos30o / 0.41 = 352.5 N

(b) Let Nf and Nb be the normal reactions on the front and back legs respectively.

For no tipping, net torque on the table about the bottom of the front legs should be zero.

=> (Fcos30o)H + NbL = W(L/2)

=> Nb = W/2 - (Fcos30o)H/L = 436/2 - (166.9 * cos30o * 0.86 / 1.75) = 147 N

Since Nb < N = 352.5 N, the table will not tip.

So,

Nf = N - Nb = 352.5 - 147 = 205.5 N

Friction on the back legs, fb = sNb = 0.41 * 147 = 60.3 N

Friction on the front legs, ff = sNf = 0.41 * 205.5 = 84.3 N

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