A symmetrical table of height 0.700 m, length 1.30 m, and weight 463 N is dragge

Question

A symmetrical table of height 0.700 m, length 1.30 m, and weight 463 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 0.405.

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

front legs:normal force

frictional force

back legs: normal force

frictional force

Explanation / Answer

here,

weight of table,w : 463 N
coefficient of friction,u :0.405

Assuming;
F applied force
Ff = frictional force which always act opposite to movement of system
N = Normal force

Wiring components of forces :
fx = f*cos30
fy = f*sin30

From newton second law :
fx = 0
fcos30 - Ff = 0
fcos30 = ff
fcos30 = uN (ff = uN) -------------(1)

Also in y direction :
N - w - fsin30 = 0
N = w + fsin30 -------------(2)

Equation 1 can be written as:
fcos30 = u(w + fsin30)

soving for minimum force, f

f(cos30-usin30) = u*w
f = u*w / (cos30-usin30)
f = (0.405*463)/(0.866 - 0.405*0.5)
f = 282.615 N

The Normal force is just force applied by floor on object so,normal force acting each leg is :

From eqn 2
N = w + fsin30
N = 463 + 282.615*0.5
N = 604.307 N

Similarly For each leg Frictional force will be same :

Ff = uN
Ff = 0.405 * 604.307
Ff = 244.744 N

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