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physics help please?
1. Two forces, 414 N at 8◦ and 341 N at 30◦ are applied to a car in an effort to accelerate it.
414 N
83175 kg---------------------------------------- 30◦ 341 NWhat is the magnitude of the resultant ofthese two forces?Answer in units of N
2. Find the direction of the resultant force (in relation to forward, with counterclockwise considered positive) Answer in degrees from the positive x-axis with counter-clockwise positive, within the limits of −180◦ to 180◦
.
Answer in units of ◦
3. If the car has a mass of 3175 kg, what magnitude of acceleration does it have?
Ignore friction.
Answer in units of m/s^2
1 Answer
- SlowfingerLv 65 months ago
Magnitude od resultant force
Fres² = F1² + F2² - 2F1F2cos𝜃
where
F1 = 414 N
F2 = 341 N
𝜃 = 30° - 8° = 22°
Fres = √(F1² + F2² + 2F1F2cos𝜃)
Fres = √(414² + 341² + 2*414*341*cos22°)
Fres = 741.3 N
Horizontal component of the resultant
Fx = F1 cos 8° + F2 cos 30°
Fx = 414 cos 8° + 341 cos 30° = 705.3 N
direction angle od the resultant
𝜷 = arccos (Fx / Fres)
𝜷 = arccos (705.3 / 741.3) = 17.9°
Only component in direction of movement Fx is relevant for acceleration.
a = Fx / m
a = 705.3 / 3175 = 0.22 m/s²
