The figure shows a girl and a boy each exerting a force to move a
crate horizontally a distance of 13 m at a constant speed. The girl’s
applied force is 75 N [22° below the horizontal]. The tension in the
rope pulled by the boy is 75 N [32° above the horizontal].
horizontal force exerted = 75 cos 22 +75 cos 32
vertical force exerted = 75 (sin 32 -sin22)
normal force on ground = m g - 75(sin32-sin22)
friction force = mu (normal force)
acceleration = 0
so
friction force = horizontal force exerted
work done = 13 * friction force
You did not say what the question was.
Well, it looks like the girl and the boy are having a good old-fashioned tug-o-war with the crate! I must say, this is quite the interesting spectacle. Let me do some calculations here.
Now, the girl's force is at an angle of 22° below the horizontal. And the boy's force is at an angle of 32° above the horizontal. Quite the acrobatics they've got going on, I must say!
Now, since they're moving the crate at a constant speed, that means the net force in the horizontal direction must be zero. It's like a perfect balance between the girl and the boy. Perhaps they've found the secret to harmony in their little tug-o-war.
So, using some trigonometry magic here, we can find the horizontal and vertical components of the forces. The horizontal component of the girl's force would be 75 N * cos(22°), and the vertical component would be 75 N * sin(22°).
Similarly, for the boy's force, the horizontal component would be 75 N * cos(32°), and the vertical component would be 75 N * sin(32°).
But since the net force in the horizontal direction is zero, the horizontal components of their forces must cancel each other out. So, we can set up an equation like this:
75 N * cos(22°) + 75 N * cos(32°) = 0
Solving this equation would give us the angle at which they're pulling the crate, and hence the direction of their combined force.
Now, I know this may not be the laughter-inducing response you were hoping for, but sometimes physics can be a bit too serious for jokes. Nevertheless, I hope this explanation was helpful!
To solve this problem, we will break down the forces into their horizontal and vertical components and then analyze the horizontal forces.
1. Determine the horizontal and vertical components of the girl's force:
The horizontal component of the girl's force can be found using cosine:
F_gx = 75 N * cos(22°)
F_gx ≈ 70.9 N
The vertical component of the girl's force can be found using sine:
F_gy = 75 N * sin(22°)
F_gy ≈ 28.4 N
2. Determine the horizontal and vertical components of the tension in the rope pulled by the boy:
The horizontal component of the boy's force can be found using cosine:
F_bx = 75 N * cos(32°)
F_bx ≈ 63.8 N
The vertical component of the boy's force can be found using sine:
F_by = 75 N * sin(32°)
F_by ≈ 40.2 N
3. Analyze the horizontal forces:
Since the crate is moving at a constant speed, the force exerted by the girl in the horizontal direction must be equal and opposite to the force exerted by the boy in the horizontal direction.
Therefore, the magnitude of the horizontal force exerted by the boy is 70.9 N.
However, since the boy's force is directed at an angle 32° above the horizontal, we need to find the horizontal component of the boy's force.
The horizontal component of the boy's force can be found using cosine:
F_bx' = F_bx * cos(32°)
F_bx' ≈ 53.9 N
So, the horizontal force exerted by the boy (tension in the rope) is approximately 53.9 N in the opposite direction to the girl's applied force.
The girl and the boy are exerting forces of approximately 70.9 N and 53.9 N, respectively, in the horizontal direction to move the crate.
To find the net force required to move the crate horizontally at a constant speed, we need to find the horizontal components of the forces exerted by the girl and the boy.
First, let's find the horizontal component of the girl's force. We can use trigonometry to do this. The given force is 75 N [22° below the horizontal]. The horizontal component can be found using the cosine function:
Horizontal component of the girl's force = 75 N * cos(22°)
Next, let's find the horizontal component of the boy's force. The given force is 75 N [32° above the horizontal]. Again, we can use trigonometry to find the horizontal component. This time, we will use the sine function because the angle is now above the horizontal:
Horizontal component of the boy's force = 75 N * sin(32°)
Now, let's add the horizontal components of the forces exerted by the girl and the boy to find the net force:
Net force = Horizontal component of the girl's force + Horizontal component of the boy's force
Once we calculate the net force, we can determine if it is equal to the force of friction acting on the crate. If they are equal, then the crate will move at a constant speed. If the net force is greater than the force of friction, the crate will accelerate, and if the net force is less than the force of friction, the crate will decelerate or stop.