A man pulls a box with a force of 70N along the positive y-axis,while a boy pulls it with a force of 50N,making an angle of 80 degrees counter clock-wise with the positive y-axis.
(i)In what direction should a second boy apply a force of 60N so that the box move along the positive y-axis?Hint;Two values are Posible.
(ii). Find the magnitude and direction of the force that a second man should apply to that the box does not move at all?
PLS HELP ME
F1 = 70N[90o].
F2 = 50N[170o], CCW from +x-axis.
F3 = 60N[Xo].
F3*Cos X = -F2*Cos170.
60*Cos X = -50*Cos170,
Cos X = 0.82067, X = 34.8o.
F1 = 70N[90o].
F2 = 50N[170o], CCW from +x-axis.
F3 = 60N[Xo].
1. The hor. component of F2 and F3 must be equal and opposite:
F3*Cos X = -F2*Cos170.
60*Cos X = -50*Cos170,
Cos X = 0.82067, X = 34.8o.
2.
i want the solutions
2. F4 =-(F1*sin90+F2*sin170+F3*sin34.8)F4 = -(70 + 8.68 + 34.2) = -113 N. = 113 N. Downward.
It is help
It's very much helpful
To solve this problem, we can break down the forces into their x-component and y-component. Since the box is being pulled along the positive y-axis, we need to find the components of each force that act in the positive y-axis direction.
(i) In what direction should a second boy apply a force of 60N so that the box moves along the positive y-axis?
To determine the direction, we need to find the angle at which the second boy should apply the force. Let's break down the given forces into their x-component and y-component.
Given:
Man's force = 70N (along the positive y-axis)
Boy's force = 50N (80 degrees counter clockwise with the positive y-axis)
Calculating the boy's force components:
Boy's x-component = 50N * cos(80°)
Boy's y-component = 50N * sin(80°)
Now, we consider the second boy's force. The box should move along the positive y-axis, so the y-components of all forces should add up to a positive value.
Let's assume the second boy applies a force at an angle θ counter clockwise with the positive y-axis and a magnitude of 60N.
For the y-components:
Positive y-component (from the man) + Boy's y-component (already calculated) + Second boy's y-component = 0
The equation becomes:
70N + 50N * sin(80°) + 60N * sin(θ) = 0
Simplifying the equation:
180N + 50N * sin(80°) + 60N * sin(θ) = 0
Now, we can solve this equation for two possible values of θ that will make the box move along the positive y-axis.
(ii) Find the magnitude and direction of the force that a second man should apply to prevent the box from moving.
To find the magnitude and direction of the force required to prevent the box from moving, we need to balance the forces in both the x and y directions.
Given:
Man's force = 70N (along the positive y-axis)
Boy's force = 50N (80 degrees counter clockwise with the positive y-axis)
Let's break down these forces into their x-component and y-component.
Calculating the man's force components:
Man's x-component = 70N * cos(90°) = 0
Man's y-component = 70N * sin(90°) = 70N
Calculating the boy's force components:
Boy's x-component = 50N * cos(80°)
Boy's y-component = 50N * sin(80°)
To prevent the box from moving, the sum of the x-components and y-components of the forces should be zero.
Equating the x-components:
0 + Boy's x-component + Second man's x-component = 0
Equating the y-components:
70N + Boy's y-component + Second man's y-component = 0
From these two equations, you can determine the magnitude and direction of the force that the second man should apply to prevent the box from moving.