While standing on a long board resting on two supports, a 70.0 kg painter paints
ID: 1701202 • Letter: W
Question
While standing on a long board resting on two supports, a 70.0 kg painter paints the side of a house. The right support is 1.50 m away from the right end of the board, and the left support is 1.50 m away from the left end of the board. The two supports are 2.50 m away from each other. The painter is 2.0 m away from the left end of the board.
A) If the mass of the board is 15.0 kg, what are the forces exerted on the two supports?
B) How close to the left end of the board can the painter move before the board tips?
Explanation / Answer
the total length of the board = 5.50 m and mid point of the board 2.75 m
If the system is equillibrium, the net force acting on the system is zero(Fnet = 0).hence torque = 0
Taking F = 0 and = 0 components,
F = 0,n1+n2 = mg + Mg -------------------- (1)
= 0 ,n1(0)+Mg(1.25)+mg(2)-n2 (2.50)= 0 ------------------- (2)
(2.50)n2 = 1.25*15*9.8 + 70*2*9.8
(2.50)n2 = 183.75 + 1372
= 1555.75
n2 = 622.3 N
substituting 'n2' in equation (1), we get
n1 = mg + Mg -622.3
= 833-622.3
= 210.7 N
Therefore the forces exerted on the two supports is 210.7 N and 622.3 N respectivly.
If suppose the painter move 'd' distance,before board is tips,hence n1 = 0 . From equation (1)
n2 = mg+Mg = 833 N
sustituting 'n2' in equation (2), we get
mg(d) = n2-Mg
d = 1m
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