Question

Tony pulls a luggage cart at an angle of 30° to the horizontal. The luggage cart has a mass of 35 kg, and he pulls with a force of 75 N.a. What is the normal force on the luggage cart?b. What force would Tony need to apply at a 30° angle to pull the luggage cart completely off the ground?c. How does your answer in part (b) change as the angle increases?INFO INCLUDED BEFORE QUESTION (picture):

Answer 1

First, draw a force diagram of the luggage cart:

Assuming that the luggage does not accelerate in the vertical direction, from Newton's Second Law of Motion we know that:

`\Sigma F_y=0`

On the other hand, the sum of the forces acting on the vertical direction is:

`\Sigma F_y=N+F\sin\theta-mg`

Then, the normal force on the luggage cart can be obtained as:

`\begin{gathered} N+F\sin\theta-mg=0 \\ \\ \Rightarrow N=mg-F\sin\theta \end{gathered}`

Part a)

To find the normal force on the luggage cart, replace m=35kg, g=9.8m/s^2, F=75N and θ=30º:

`\begin{gathered} N=mg-F\sin\theta \\ \\ =(35kg)(9.8(m)/(s^2))-(75N)(\sin(30º)) \\ \\ =343N-37.5N \\ \\ =305.5N \end{gathered}`

Therefore, the normal force on the luggage cart is 305.5N.

Part b)

When the luggage cart is pulled completely off the ground, the normal force becomes equal to 0. Then:

`\begin{gathered} 0=mg-F\sin\theta \\ \\ \Rightarrow F=(mg)/(\sin\theta) \end{gathered}`

Replace the values of the known variables to find the magnitude F:

`F=((35kg)(9.8(m)/(s^2)))/(\sin(30º))=686N`

Therefore, Tony would need to apply a force of 686N to pull the luggage cart completely off the ground.

Part c)

If the angle increases, then the sine of that angle also increases. Then, the denominator of the expression grows, causing the full expression to shrink.

Therefore, if the angle increases, then the force needed to pull the luggage completely off the ground decreases, following the relation:

`F=(mg)/(\sin\theta)`