Question

A wedge has to be driven for tightening a body angle 15 degree loaded with 1000N force is shown in figure. If coefficient of friction is 0.25 for all surfaces. Determine the force, which should be applied to the widge.

Answer 1

Answer:

Force that should apply to the wedge is 16.8 N

Explanation:

Let the normal force due to side wall is N1 and due to lower wedge it is N2

so we will have

$N_1 cos15 + \mu N_2 = 1000 sin15$

$N_2 = N_1 sin15 + 1000 cos15$

now from above equations

$N_1 cos15 + \mu (N_1 sin15 + 1000 cos15) = 1000 sin15$

$N_1 (cos15 + 0.25 sin15) = 1000(sin15 - 0.25 cos15)$

$N_1 = 16.8 N$

$N_2 = 970.3 N$

Now on lower wedge we have

$P + F_f cos15  = N_2 sin15$

$P = N_2 sin15 - F_f cos15$

$P = N_2 sin15 - \mu N_2 cos15$

$P = 970.3(sin15 - 0.25 cos15)$

$P = 16.8 N$

#Learn

Topic : Friction force

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