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°counter clockwise with the positive Y-axis.
(1)In what direction should the second boy apply a force of 60N so that the box will move along the positive Y-axis? Hint;two values are possible
F1 = 70[90o] CCW from +x-axis
F2 = 50[170o] CCW from +x-axis.
F3 = 60[Ao] from +x-axis.
Sum of hor. components = 0:
70*Cos90 + 50*Cos170 + 60*Cos A = 0.
60*Cos A = 49.2.
A = 34.9 degrees.
well,
along the x axis: You want the net force in that direction to be zero.
70Sin0deg+50sin80+60sinTheta=0
sinTheta=-50sin80deg/60= -0.820673128
Theta=-(55.1522333 deg) or -(-55.1522333 deg+180deg) or -124.8deg
50*sin280 = -60*sin A,
- 49.24 = -60*sin A,
sin A = 49.2/60,
A = 55.1o CW from +Y-axis = 34.9o CCW from +X-axis.
To find the direction in which the second boy should apply a force of 60N so that the box moves along the positive Y-axis, we need to consider the forces acting on the box and their resultant.
Given:
Force applied by the man (F₁) = 70N along the positive Y-axis.
Force applied by the first boy (F₂) = 50N at an angle of 80° counter clockwise with the positive Y-axis.
To determine the direction in which the second boy should apply a force, we need to balance the forces in the Y-direction.
Step 1: Resolve the forces into their respective components.
The vertical component of F₁ = F₁ * sin(0°) = F₁ * 1 = 70N * 1 = 70N along the positive Y-axis.
The horizontal component of F₁ = F₁ * cos(0°) = F₁ * 0 = 70N * 0 = 0N along the X-axis.
The vertical component of F₂ = F₂ * sin(80°) = 50N * sin(80°).
The horizontal component of F₂ = F₂ * cos(80°) = 50N * cos(80°).
Step 2: Calculate the resultant vertical force.
The resultant vertical force (R) is the sum of the vertical components of the forces applied by the man and the first boy.
R = F₁ * sin(0°) + F₂ * sin(80°)
Step 3: Set up an equation to solve for the direction of the second boy's force.
To move the box along the positive Y-axis, the resultant vertical force should be equal to the force applied by the second boy.
R = F₃ * sin(θ)
where F₃ = 60N (force applied by the second boy) and θ is the angle between the force applied by the second boy and the positive Y-axis.
So, we have: R = 60N * sin(θ)
Setting up the equation:
60N * sin(θ) = F₁ * sin(0°) + F₂ * sin(80°)
Now, plug in the values and solve for θ.
60N * sin(θ) = 70N * 1 + 50N * sin(80°)
Simplifying:
60N * sin(θ) = 70N + 50N * sin(80°)
Divide both sides by 60N:
sin(θ) = (70N + 50N * sin(80°))/60N
Now, using a scientific calculator or trigonometric table, find the two possible values of θ that satisfy this equation. This will give you the two possible directions in which the second boy should apply a force of 60N so that the box will move along the positive Y-axis.