a 882N man is 1/4 of the way up a 10m ladder that is resting on a smooth frictionless wall. if the mass of the ladder is 343N and the ladder makes an angle of 60deg with the ground, find the force of friction of the ground on the foot of the ladder. Also find the vertical force at point of contact with the ground.
2007-03-22 04:21:01 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The vertical force in contact with the ground is equal to the sum of the weights of the ladder and man since it is the only vertical reactive force.
=343+882
=1225 N
The frictional force is found by summing the torques at the point of contact with the wall.
2007-03-22 05:45:00 · answer #1 · answered by odu83 7 · 0⤊ 0⤋
the system is in rotational and static equilibrum
. The vectorial sum of all the forces on the ladder is 0 and the sum of all torque is 0
On Y axis we have (Reference point A on the ground)
N-882-343=0
on X axis we have
Fb(x)-ff=0
where Fb(x) is reaction of the wall at point of contact and ff is the friction force. Those are the only two forces acting horizontally on the ladder.
we have to fin Fb(x) and it is found by considering that the sum of all torque around point A ( contact with ground) is 0
The simplest definition of torque To is
To= L*F( perpendicular to L) or To= F*L (eff)
If the point of contact with the ground is the point of rotation.The sum of torque is 0
then
sum ot To = 882*2.5*cos 60 +343*5*cos60-Fb(x)*10 sin 60=0
then Fb(X)= cos60(882*2.5+343*5)/10sin60
with Fb(x)=ff ....it is easy to find ff
If I was your physics teacher I would have asked you an other question....
what is the static coefficent of friction betwwen the ground and the ladder
P.A
2007-03-22 08:03:30 · answer #2 · answered by pieall2003 3 · 0⤊ 0⤋
Ff = (882 + 343)sin60°cos60° = 530.44 N
Fn = (882 + 343) = 1225 N
2007-03-22 07:20:11 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋