Please help with this problem. My Notes 6. 24 points | Previous Answers OSColPhy

Question

Please help with this problem.

My Notes 6. 24 points | Previous Answers OSColPhys2016 4.5.WA.035. Tom enlists the help of his friend John to move his car. They apply forces to the car as shown in the diagram. Here F1436 N and F2 389 N and friction is negligible. In the diagram below, the mass of the car 3500 kg, 6125 and 2-12°, Assume the car faces the positive x-axis before the forces are applied.) ' (a) Find the resultant force exerted on the car. magnitude 775.65 17.4 The question asks you to find the overall force exerted on the car, not just in the x or y direction. (counterclockwise from the +x-axis) direction (b) What is the acceleration of the car? 1861 magnitude How is the net force on an object related to its acceleration? m/s direction (counterclockwise from the +x-axis) Additional Materials

Explanation / Answer

given,F1 = 436 N

F2 = 389 N

theta 2 = 12 deg

theta 1 = -25 deg

mass of car m = 3500 kg

a. considering i and j to be unit vectors along x and y direction

Resultant force = F1(cos(theta1)i + sin(theta1)j) + F2(cos(theta2)i + sin(theta2)j)

F = 436(cos(12)i + sin(12j) + 389(cos(25)i - sin(25)j)) = 779.026i -73.749 j

so magnitude = sqroot[(779.026)^2 + (73.749)^2) = 782.509 N

angle = arctan(-73.749/779.026) = -5.4079 deg

b. acceleration of car = F/m

magnitude = 782.509/3500 = 0.223574 m/s/s

directoin = -5.4079 deg

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---