A father gives his son a ride by pulling a sled across the rough snow. The father runs across the snow with a constant acceleration while pulling the sled at F = 75.0 N, at 30 deg. above the horizontal. The combined mass the sled and son is 30.0 kg. The coefficient of kinetic friction between the sled and the snow is Uk = 0.200.
a) What is the magnitude of the normal force exerted on the sled-son system?
sum of Fx: -fk + 75cos30 = max
sum of Fy: n-mg +75sin30 = may
This is as far as I got and I'm not even sure if it's correct. Please provide a concise solution so that I can see exactly what's going on. I need to master Newton's laws. I have pressure of college approaching. If I see the way, I can solve problems. Also, I don't just want solution as not knowing what is happening won't help me on a test. I have tried to solve it on paper. It is tedious to put my inaccurate solution on here. Thank you so much.
b) What is the magnitude of the acceleration of the sled-son system?
I may have figured it out: Here's the solution.
sum Fx: 75cos30 - 0.200 = max
sum Fy: n-mg + 75 sin30 = 0
n = mg - 75 sin30
n = (30*9.8) - 75sin30
n = 257 N
b) a = (75 cos 30 - 0.200) / 30kg
a = 2.2 m/s^2
check my thinking.
Your thinking for part a is correct. Let's go through the solution step by step to understand the problem better:
For part a, we need to find the magnitude of the normal force exerted on the sled-son system. We can start by analyzing the forces acting on the system in the horizontal and vertical directions.
In the horizontal direction, the only force acting is the force of kinetic friction (fk), which is given by the coefficient of kinetic friction (μk) multiplied by the normal force (n). Note that the normal force acts perpendicular to the surface of the snow and counterbalances the weight of the system (mg).
For the horizontal direction, using the equation of motion, we have:
sum of Fx: -fk + 75cos30 = max
Now, let's determine the value of fk. Since the system is moving, we can use the equation fk = μk * n, where μk is given as 0.200. Substituting this into our equation:
-0.200n + 75cos30 = max
In the vertical direction, we have two forces acting: the normal force (n) and the vertical component of the pulling force (75sin30) pointing upwards. These forces need to balance the weight of the system (mg).
For the vertical direction, using the equation of motion, we have:
sum of Fy: n - mg + 75sin30 = may
Let's substitute the value of n from the horizontal equation into this equation:
(mg - 75sin30) - mg + 75sin30 = may
-mg + mg - 75sin30 + 75sin30 = may
0 = may
From this equation, we can see that the acceleration in the vertical direction (ay) is zero because there is no net force acting in that direction. This means that the normal force is equal to the weight of the system:
n = mg
Substituting the given values:
n = (30 kg * 9.8 m/s^2)
n = 294 N
So the magnitude of the normal force exerted on the sled-son system is 294 N.
For part b, we need to find the magnitude of the acceleration of the sled-son system. We can use the horizontal equation (sum of Fx) to find the acceleration (a):
75cos30 - 0.200n = max
Substituting the value of n (294 N) into this equation:
75cos30 - 0.200 * 294 = 30a
30a = 2.2
a = 2.2 m/s^2
Thus, the magnitude of the acceleration of the sled-son system is 2.2 m/s^2.
In summary:
a) The magnitude of the normal force exerted on the sled-son system is 294 N.
b) The magnitude of the acceleration of the sled-son system is 2.2 m/s^2.