A 10m-ladder is placed against a wall. The coe?cient of friction is 0.45 between

Question

A 10m-ladder is placed against a wall. The coe?cient of friction is 0.45 between the ladder and the wall and also between the ladder and the ?oor, both of which are glossy surfaces. A 70kg-man places this 5kg-ladder at an angle 60? with respect to the ground and starts climbing the ladder. Write the equation for the sum of the torques on the ladder. Solve your equations to ?nd how high up the ladder the man can go without the ladder slipping. You can solve for either the vertical distance or the distance along the ladder.

Explanation / Answer

Vertical Force F= F(ladder) + F(man)

= (5Kg)(9.8m/s) + (45Kg)((9.8m/s)

= 49N + 441N

=490N

Max Frictional Force = (.45)(490N)

=220.5N

Normal force: = Vertical force

Angle made by ladder with wall:

?=90-60 =30

Let distance climbed by man is x.

By balancing torque about wall , we have

49*10/2(cos(30)+441*(10-x)*cos(30)+220.5*10*sin(30)-490*10*cos(30)=0

10-x = 7.66

x=2.34m

Distance travelled along the ladder = 2.34m

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