# Problem 914 | Combined Axial and Bending

anil

a block weighting 1000N is kept on a rough plane inclines at 40 digress to the horizontal the coefficient of friction between the block and the plane is 0.6.Determine the smallest force inclines at 15 digress to the plane required just to move the block up the plane.

Neil tambz

Let: P=force needed to move up the block

Take summation of forces vertical
Fv=0;

N=1000cos40+Psin15 eq.1

Take summation of forces horizontal Fh=0;

Pcos15=F+100sin40 eq.2

note: F=uN

where F stands for friction force
u stands for coefficient of friction
N stands for normal force on plane

Subst. eq. 1 to the formula;

F=0.6[(1000cos40+Psin15] eq.3

Then subst. eq.3 to eq.2;

Pcos15=0.6(1000cos40+Psin15)+1000sin40

Solving for P;

P=1359.94 N answer

Hope this will help you..

Mithran Venkatesh sir what software you are using to draw the beam diagram with different support conditions like pin and dollar supports

MATHalino MS Visio for figures in this page.

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