A 2.6 m long, 16 kg piece of plywood is sitting on a pair of sawhorses. The sawhorses are .3 m from the left end and .5 m in from the right end of the plywood. A 4 kg circular saw is sitting 1 m in from the left end of the plywood. How much force is each sawhorse applying? Assume the plywood is of uniform constant.
Again, I'm lost. Any help is much appreciated.
A 2.6 m long, 16 kg piece of plywood is sitting on a pair of sawhorses. The sawhorses are .3 m from the left end and .5 m in from the right end of the plywood. A 4 kg circular saw is sitting 1 m in from the left end of the plywood. How much force is each sawhorse applying? Assume the plywood is of uniform constant.
.............6.1538kg/m
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....../\.....\/.........../\
..3m..Rl.7m.4kg...1.8m....Rr.5m
Rl and Rr are the left and right reactions respectively.
Taking the summation of moments about the left reaction, we have
-.3(6.1538).15 + 4(.7) + 2.3(6.1538)1.15 - 1.8rR.
Solve for Rr.
Take the summation of vertical forces to derive the left reaction.
......RL................Rr
To determine the force each sawhorse is applying, we can start by calculating the center of mass of the plywood.
First, let's find the total mass of the plywood and circular saw:
Total Mass = Mass of Plywood + Mass of Circular Saw
Total Mass = 16 kg + 4 kg
Total Mass = 20 kg
Next, we can find the position of the center of mass. Since the plywood is uniform, the center of mass will be at the midpoint of its length:
Center of Mass Position = (Length of Plywood) / 2
Center of Mass Position = 2.6 m / 2
Center of Mass Position = 1.3 m
Now, we can calculate the torque on the plywood about the left end (where the sawhorses are located).
Torque = Force * Distance
Torque due to left sawhorse = Force of left sawhorse * Distance from left end
Torque due to right sawhorse = Force of right sawhorse * Distance from left end
The sum of these torques should be zero, since there is no rotational acceleration:
Torque due to left sawhorse + Torque due to right sawhorse = 0
Since the center of mass is the pivot point, the torque due to the center of mass is also zero:
Torque due to center of mass = 0
To calculate the torque, we multiply the force exerted by each sawhorse by its respective distance from the left end of the plywood.
We'll use x to represent the force of the left sawhorse and y to represent the force of the right sawhorse.
x * (0.3 m) + y * (2.1 m) = 0
Now, we have one equation with two unknowns. However, we can solve this system of equations using another principle: the vertical forces on the plywood must balance out.
Vertical Force up = Vertical Force down
The vertical forces on the plywood are the gravitational forces acting on each component (plywood and circular saw).
Vertical force up = Force of Left Sawhorse + Force of Right Sawhorse + Weight of Plywood + Weight of Circular Saw
Vertical force down = Total Mass * gravitational acceleration
Plugging in the given values:
x + y + 16 kg * 9.8 m/s^2 + 4 kg * 9.8 m/s^2 = 20 kg * 9.8 m/s^2
Simplifying the equation:
x + y + 156.8 N + 39.2 N = 196 N
x + y = 0.8 N
Now, we have a system of two equations with two unknowns:
x * (0.3 m) + y * (2.1 m) = 0
x + y = 0.8 N
We can solve this system of equations to find the values of x and y, representing the forces applied by the left and right sawhorses, respectively.
Solving this system, we find:
x = 0.6 N
y = 0.2 N
Hence, the left sawhorse is applying a force of 0.6 N, and the right sawhorse is applying a force of 0.2 N.