A crate sets on wooden horizontal surface (a wooden floor). The figure shows a top view of this looking down onto the crate (gravity would be acting into the page). Man one and man two apply forces F1 and F2, at angles of θ1 and θ2 respectively, with the goal of moving the crate in the x-direction. A resultant force of Fr = 29 lbs in the x-direction is required to accomplish this. All of the forces are in the xy plane. If man one applies a force of F1 = 19.5 lbs at an angle of θ1 = 17° from the positive x-axis, complete the following steps to determine the magnitude and angle of the force man two must apply.

Question

A crate sets on wooden horizontal surface (a wooden floor). The figure shows a top view of this looking down onto the crate (gravity would be acting into the page). Man one and man two apply forces F1 and F2, at angles of θ1 and θ2 respectively, with the goal of moving the crate in the x-direction. A resultant force of Fr = 29 lbs in the x-direction is required to accomplish this. All of the forces are in the xy plane. If man one applies a force of F1 = 19.5 lbs at an angle of θ1 = 17° from the positive x-axis, complete the following steps to determine the magnitude and angle of the force man two must apply.

Fr = 29 lbs
F1 = 19.5 lbs
θ1 = 17°

A. Write an expression for Fr, just as the crate starting to move using the sum of the factors in the x direction in terms of the variables F1, F2, θ1, and θ2.

B.Write an equation for the sum of forces in the y direction when the crate just starts to move using the specified coordinate system. Put your answer in terms of the variables F1, F2, θ1, and θ2.

C. Combine these two equations to develop an expression for tan(θ2) in terms of Fr, F1, F2, and θ1. Remember that the crate does not move along the y-direction.

D. Solve numerically for the value of θ2 in degrees.

E. Using this value for θ2 and other known values, solve numerically for the value of F2 in lbs.

Answer 1

A. To determine the expression for Fr, we need to consider the x-components of the forces F1 and F2. The x-component of F1 is F1x = F1 * cos(θ1), and the x-component of F2 is F2x = F2 * cos(θ2). Therefore, the expression for Fr is:

Fr = F1x + F2x

B. Since the crate is on a horizontal surface and not moving vertically, the sum of forces in the y-direction should be zero. The y-component of F1 is F1y = F1 * sin(θ1), and the y-component of F2 is F2y = F2 * sin(θ2). Therefore, the equation for the sum of forces in the y-direction is:

F1y + F2y = 0

C. Combining the expressions from parts A and B, we have:

F1x + F2x = Fr
F1 * cos(θ1) + F2 * cos(θ2) = Fr

Since we know that Fr = 29 lbs, and F1 = 19.5 lbs and θ1 = 17°, we can substitute these values into the equation and solve for tan(θ2):

19.5 * cos(17°) + F2 * cos(θ2) = 29
F2 * cos(θ2) = 29 - 19.5 * cos(17°)
tan(θ2) = (29 - 19.5 * cos(17°)) / F2

D. To solve numerically for θ2, we need to know the value of F2.

E. Once we have the value of θ2 from part D, we can substitute it into the equation from part C to solve for F2:

tan(θ2) = (29 - 19.5 * cos(17°)) / F2

By rearranging the equation, we can solve for F2:

F2 = (29 - 19.5 * cos(17°)) / tan(θ2)

Again, we need the numerical value of θ2 to calculate F2.

Answer 2

A. To write an expression for Fr in terms of F1, F2, θ1, and θ2, we can use the principle of vector addition. Since the crate is moving in the x-direction, the net force in the x-direction should be equal to Fr.

The x-component of F1 is F1x = F1 * cos(θ1), and the x-component of F2 is F2x = F2 * cos(θ2).

So, the expression for Fr in the x-direction would be: Frx = F1x + F2x = F1 * cos(θ1) + F2 * cos(θ2).

B. In the y-direction, since the crate is not moving vertically, the sum of the forces in the y-direction should be zero.

The y-component of F1 is F1y = F1 * sin(θ1), and the y-component of F2 is F2y = F2 * sin(θ2).

So, the equation for the sum of forces in the y-direction would be: F1y + F2y = F1 * sin(θ1) + F2 * sin(θ2) = 0.

C. Combining the two equations from A and B, we get:

F1 * cos(θ1) + F2 * cos(θ2) = Fr .........(1)

F1 * sin(θ1) + F2 * sin(θ2) = 0 ..........(2)

Now, let's solve equation (2) for F2 * sin(θ2):

F1 * sin(θ1) + F2 * sin(θ2) = 0

F2 * sin(θ2) = -F1 * sin(θ1)

Dividing both sides by sin(θ2), we get:

F2 = (-F1 * sin(θ1)) / sin(θ2)

Now, let's solve equation (1) for cos(θ2):

F1 * cos(θ1) + F2 * cos(θ2) = Fr

F2 * cos(θ2) = Fr - F1 * cos(θ1)

Dividing both sides by cos(θ2), we get:

F2 = (Fr - F1 * cos(θ1)) / cos(θ2)

D. Now, we have the expression for F2 in terms of known quantities. To solve numerically for θ2, we need to substitute the given values for Fr, F1, and θ1 into the equation above and solve for θ2.

E. Once we have the value of θ2, we can substitute it back into the expression for F2 to solve numerically for the value of F2. Again, we need to substitute the known values for Fr, F1, θ1, and θ2 into the equation for F2 and solve for F2.

no

A. To determine the expression for Fr, we need to consider the x-components of the forces F1 and F2. The x-component of F1 is F1x = F1 * cos(θ1), and the x-component of F2 is F2x = F2 * cos(θ2). Therefore, the expression for Fr is:

A. To write an expression for Fr in terms of F1, F2, θ1, and θ2, we can use the principle of vector addition. Since the crate is moving in the x-direction, the net force in the x-direction should be equal to Fr.

NO