1- Two forces are applied to a car in an effort to move it, as shown in the foll
ID: 2275375 • Letter: 1
Question
1- Two forces are applied to a car in an effort to move it, as shown in the following figure, where F1 = 433 N and F2 = 373 N. (Assume up and to the right as positive directions.)
Two forces are applied to a car in an effort to move it, as shown in the following figure, where F1 = 433 N and F2 = 373 N. (Assume up and to the right as positive directions.) What is the resultant of these two forces? magnitude = ? N direction = ? degree to the right of the forward direction If the car has a mass of 3,000 kg, what acceleration does it have? Ignore friction. = ? m/s2 2- wo horizontal forces, and , are acting on a box, but only is shown in the drawing. can point either to the right or to the left. The box moves only along the x axis. There is no friction between the box and the surface. Suppose that = +6.0 N and the mass of the box is 2.2 kg. What is the algebraic expression that gives the magnitude and direction of ? Note that the direction to the right is the positive direction in the drawing. Express your answer in terms of the mass m of the block, the acceleration ax, and the force F1? Find the magnitude and direction of when the acceleration of the box is +4.0 m/s2. Specify the direction by including in your answer an explicit algebraic sign along with the magnitude. Find the magnitude and direction of when the acceleration of the box is -4.5 m/s2. Specify the direction by including in your answer an explicit algebraic sign along with the magnitude. Find the magnitude and direction of when the acceleration of the box is 0 m/s2. Specify the direction by including in your answer an explicit algebraic sign along with the magnitude.Explanation / Answer
a . Referring to the figure....
Resolving the forces we ve
F1 = 433 cos10 i + 433 sin10 j
= 426.42i + 75.19 j
F2 = 373 cos30 i - 373 sin30 j
= 323.03 i - 186.5j
Hnece net force ll be F net = 426.42i + 75.19 j + 323.03 i - 186.5j
= 749.45 i - 111.31j
Magnitude ll be = sqrt (749.45^2 + 111.31^2) = 757.671 N
Direction theta = tan inverse (111.31/749.45 ) = 8.45 degree
b. mass = 3000 kg
as we know dat Force = mass* accleration
hence
acceleration = force/mass = 757.671/3000 = 0.2526 m/s^2 which is to the right of car centre line
Coming to the 2nd question we ve
a . as we know dat NET FORCE = Sumaation of all the forces
hence
Fnet = F1 + F2
in the 1st case
F2 = m*a - F1
where a is the accleration along x direction......
b. Fnet = F1 + F2
2.2*4 = 6 + F2 =
hence F2 = 2.8 N
c . Fnet = F1 + F2
2.2*(-4.5) = 6 + F2
F2 = -15.9 N
d . Fnet = F1 + F2
2.2*0 = 6+ F2
hence F2 = -6 N
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