Answer:
3. 1294 lb
4. 126 N
Step-by-step explanation:
You want the force necessary to hold a 5000 lb equipment stationary on a 15° incline, and you want the magnitude of the sum of 56 N and 76 N forces separated by 35°.
3. Incline
The down-ramp component of the weight is ...
(5000 lb)·sin(15)° ≈ 1294 lb
It takes a force of 1294 lb to hold the equipment in place on a frictionless incline at 15° from the horizontal.
(A force equal to the full weight is required if the ramp is 90°: sin(90°) = 1; and no force at all is required if the ramp is horizontal: sin(0°) = 0.)
4. Resultant
The Law of Cosines can be useful for finding the magnitude of the sum of forces. When they are added (nose to tail), the angle between them is the supplement of 35°: 145°. Then we have ...
c² = a² +b² -2ab·cos(C)
c² = 56² +76² -2·56·76·cos(145°) ≈ 15884.6
c = √(15884.6) ≈ 126.03
The two forces sum to one with a magnitude of 126 N.
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Additional comment
The "equilibrant" of a collection of forces is the opposite of their sum. It will have the same magnitude, but a direction 180° from that of the sum.
Here, the force necessary for equilibrium will have a magnitude of 126 N and a direction of about 160° from the 56 N force, away from the 76 N force.
