Suppose a crate of mass 5.6 kg is placed on the plank in the figure below at a d

Question

Suppose a crate of mass 5.6 kg is placed on the plank in the figure below at a distance x = 8.3 m from the left end. If the plank has a mass of 12 kg, find the forces exerted by the two supports on the plank. (Indicate the direction with the sign of your answer.)

a- left support (N)

b- right support (N)

Suppose a crate of mass 5.6 kg is placed on the plank in the figure below at a distance x = 8.3 m from the left end. If the plank has a mass of 12 kg, find the forces exerted by the two supports on the plank. (Indicate the direction with the sign of your answer.) a- left support (N) b- right support (N)

Explanation / Answer

This problem requires two equation since it has two unknowns. The best way to solve this problem is by using the following equations:
?M = 0; summation of force about a point = 0 under static conditions
?Fy = 0; summation of forces in the vertical direction = 0 under static conditions
To solve this problem first you need to determine the weight of the plank per meter. The plank is 12m long and has a mass of 12kg. This makes the plank's mass = 1kg / m.
Now set your zero coordinate at the left fulcum = f1; the right fulcum will be = f2. If you do a free body diagram you will have the following:
m1 = 3kg located at a point that is 1.5m to the left of f1
m2 = 7kg located at a point that is 3.5m to the right of f1
m3 = 7kg located at a point that is 5.6m to the right of f1
m4 = 2kg located at a point that is 8m to the right of f1
and the moment about f1 by f2 is equal to f2 * 7m
If you do a summation of the moment about f1 you will have (Note: Clockwise is positive):
?M = 0 = -1.5m * 3kg + 3.5m * 7kg + 5.6m * 7.2kg + 8m * 2 kg - 7m * f2; solve for f2. This gives:
f2 = [-1.5m * 3kg + 3.5m * 7kg + 5.6m * 7.2kg + 8m * 2 kg] / 7m
f2 = (-4.5m-kg + 24.5m-kg + 40.32m-kg + 16kg-m) / 7m
f2 = 10.903 kg = force exerted by the right fulcum

Now that you know f2, you can use the vertical force equation to solve for f1 as follows
?Fy = 0 = 3kg + 7kg + 7.2kg + 2kg - (F1 + F2); F2 = 10.903kg, solve for F1. This gives:
F1 = 19.2kg - 10.903kg = 8.297kg = the force exerted by the left fulcum

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