A boy drags a 100 N sled up a 20 degree slope at constant velocity. If the coefficient of friction between the sled and hill is 0.2 what force must he exert at an angle of 35 degrees with respect to the hill?
Fs = 100N @ 20 deg.
Fp = 100sin20 = 34.2N. = Force parallel
to the hill.
Fv = 100cos20 = 94N = Force perpendicular to the hill.
Ff = 0.2*94 = 18.8N = Force of friction.
Fap*cos35 - Fp - Ff = 0,
Fap*cos35 - 34.2 - 18.8 = 0,
Fap*cos35 - 53 = 0,
Fap*cos35 = 53,
Fap = 64.7N = Applied force.
Henry, you didn't account for the vertical pull of the boy. therefore friction is less than 18.8.
To find the force that the boy must exert at a 35 degree angle with respect to the hill, we first need to calculate the total force acting on the sled.
1. Calculate the vertical component of the gravitational force:
- The total weight of the sled is 100 N.
- The weight can be split into two components: one acting perpendicular to the slope and the other parallel to the slope.
- The component perpendicular to the slope is given by: 100 N * cos(20°).
So, the vertical component of the gravitational force is: 100 N * cos(20°).
2. Calculate the frictional force:
- The coefficient of friction is given as 0.2.
- The normal force acting on the sled is the weight component perpendicular to the slope.
- The frictional force is given by: coefficient of friction * normal force.
So, the frictional force is: 0.2 * (100 N * cos(20°)).
3. The net force acting on the sled must be zero since it is moving at a constant velocity. This net force is a combination of the force exerted by the boy and the frictional force.
- Let F be the force exerted by the boy.
- In the direction perpendicular to the slope, the net force is given by: F * sin(35°) - (100 N * cos(20°)).
- In the direction parallel to the slope, the net force is given by: F * cos(35°) - (0.2 * (100 N * cos(20°))).
So, the net force in the perpendicular direction is: F * sin(35°) - (100 N * cos(20°)).
And the net force in the parallel direction is: F * cos(35°) - (0.2 * (100 N * cos(20°))).
4. Since the sled is moving at constant velocity, the net force is zero.
- Set the net force in the perpendicular direction equal to zero: F * sin(35°) - (100 N * cos(20°)) = 0.
- Solve for F.
By rearranging the equation and solving for F, we can find the force that the boy must exert at a 35 degree angle with respect to the hill.
To find the force the boy must exert at an angle of 35 degrees with respect to the hill, we need to break down the forces acting on the sled and use trigonometry.
First, let's identify the forces involved:
1. The force of gravity (weight) acting vertically downwards = 100 N.
2. The normal force acting perpendicular to the hill.
3. The force of friction acting parallel to the hill.
4. The force the boy exerts at an angle of 35 degrees.
Since the sled is being dragged up the slope at a constant velocity, we know that the net force acting on the sled is zero. Therefore, the force the boy exerts should balance out the forces of friction and gravity.
Let's break down the forces into their components:
1. The force of gravity can be divided into two components:
- The component acting parallel to the hill = mg * sin(20°)
- The component acting perpendicular to the hill = mg * cos(20°)
2. The force the boy exerts can be divided into two components:
- The component acting parallel to the hill = F * cos(35°)
- The component acting perpendicular to the hill = F * sin(35°)
Now, let's write the equation for the net force:
Net force = Force uphill - Force downhill
Since the sled is at a constant velocity, the forces uphill and downhill will be equal in magnitude but opposite in direction.
Force uphill = force due to the component parallel to the hill - force due to friction
= F * cos(35°) - (Coefficient of friction * normal force)
Force downhill = force due to the component parallel to the hill + force due to gravity parallel to the hill
= F * cos(35°) + mg * sin(20°)
Since the net force is zero, we can set Force uphill equal to Force downhill and solve for F:
F * cos(35°) - (Coefficient of friction * normal force) = F * cos(35°) + mg * sin(20°)
We can replace the normal force with mg * cos(20°):
F * cos(35°) - (Coefficient of friction * mg * cos(20°)) = F * cos(35°) + mg * sin(20°)
Now, let's solve this equation for F:
F * cos(35°) - F * cos(35°) = mg * sin(20°) + (Coefficient of friction * mg * cos(20°))
0 = mg * sin(20°) + (Coefficient of friction * mg * cos(20°))
Now, we can solve for F:
F = [mg * sin(20°)] / [cos(35°) - Coefficient of friction * cos(20°)]
By plugging in the given values, we can calculate the force the boy must exert at an angle of 35 degrees with respect to the hill.