Question

Determine the magnitude of the gripping forces exerted along line aa on the nut when two 70-lb forces are applied to the handles as shown. Assume that pins A and Dslide freely in slots cut in the jaws. P 0.75 in. а -0.5 in 4.5 in.- В A E P The magnitude of the gripping force exerted along the line aa is Ib.

Answer 1

The answer is 490 lb.

The solution is given below -

Solution - 0.5in free body diagram of upper Jaw- a Let Vertical reaction at A = Ay Horizontal reaction at B = BX and Vertical reaction at B = By as shown in figure a from equilibrium Conditions. = { fx=0 Bx=0 i EMB=0 9 (0.5) - Ay 11.5 =0 3 - Ay Now, EMA=0 - By (1:5) + f $(3) = 0 By = 20 = 40

Now, free body diagram of lower Jaw- 'Ey ! Led Vertical reaction at D be by . Horizontal reaction at Ebe Ex & Vertical sin reaction and E be Ey as shown in figure. & Applying equilibrium equations- & fx = 0 y | Ex=ol I E MD=0 I Ex (1-5)-0(2) = 0 Now, Tolbl free Handle Body diagram of pronsin 4.5in. En=0 y X Ey 5.25 in.

from above Solutions lequations), А, е , су = 9 c=0 Moment at E Mc = 0 - a (70) [5.25) + Ay 10.75) - Ey (0.75)=0 367.5 + 0) 10-75) - 140 10.75) = 0 3. 367.5 - 8 10.75)=0 10 = 490 lb executed The magnitude of gripping force along the line aain 490 lb.