A construction supervisor of mass M = 83.7 kg is standing on a board of mass m = 30.5 kg. The board is supported by two saw horses that are a distance of 4.50 m apart. If the man stands a distance x1 = 1.30 m away from saw horse 1 as shown, what is the force that the board exerts on that saw horse?
Take moments about the other saw horse (#2) and set the total moment equal to zero. Let F be the upward force exerted by saw horse 1.
83.7g * 3.20 + 30.5g * 2.25 - F*4.50 = 0
Solve for F, in Newtons. g is the acceleration of gravity, 9.8 m/s^2
3.2 m is how far the man is from sawhorse 1 and 2.25 m is the distance of the center of mass of the board from sawhorse 1
its cool the answer is 732.57 N
i got it =]
this was what i did and it worked for me
ok but the Fb*d the d=0 so that cancels out so i have
83.7(-9.8)(3.2)+30.5(-9.8)(2.25)-Fa(4.5)
where the guy weighs 83.7 Kg
the board weights 30.5 kg
the distance between the saw horses is 4.5 m
and Fb of the second saw horse cancels out to 0 because the distance from the pivot point is non-existent
1. Suppose a freely orienting chain with 1000 segments each of length 7 AO is subjected to a force on its ends of 10-5 N. What will be the average separation of the chain ends?
2. How large a force is needed to elongate the following piece of polymer to a length of 2.54 cm?
Original length = 10 cm.
Area of cross section = 0.1 cm2
Mn = 30,000, Mc = 6000
Density = 0.90 gm/cc
3. Consider a polymer for which the potential energy resulting from attractive forces between the chains decreases markedly as the chains are oriented. What can one say about the effect of these forces upon the equilibrium stress-strain curve for that polymer rubber?
To find the force that the board exerts on saw horse 1, we need to consider the torque (or moment) equilibrium of the system.
Torque is the rotational force applied to an object, and it depends on the distance between the force and the pivot point. In this case, the pivot point is saw horse 2.
First, let's calculate the torque exerted by the supervisor's weight:
Torque_supervisor = (Weight of supervisor) * (Distance between supervisor's position and saw horse 2)
= M * g * x1
where M is the mass of the supervisor, g is the acceleration due to gravity, and x1 is the distance between saw horse 1 and the supervisor.
Next, let's calculate the torque exerted by the board's weight:
Torque_board = (Weight of board) * (Distance between board's center of mass and saw horse 2)
= m * g * (x1 + L/2)
where m is the mass of the board, L is the length of the board, and x1 is the distance between saw horse 1 and the supervisor.
Now, we can find the total torque exerted on saw horse 2:
Total torque = Torque_supervisor + Torque_board
Since the saw horses are symmetrically placed, the total torque exerted must be zero for equilibrium. Therefore:
Total torque = 0
Torque_supervisor + Torque_board = 0
(M * g * x1) + (m * g * (x1 + L/2)) = 0
Now, we can solve for the force that the board exerts on saw horse 1:
Force_board_on_saw_horse_1 = (Torque_board) / (Distance between saw horse 1 and saw horse 2)
Force_board_on_saw_horse_1 = (m * g * (x1 + L/2)) / (D)
where D is the distance between saw horse 1 and saw horse 2.
Substituting the given values, we can find the force that the board exerts on saw horse 1.