Question

A flat-topped toy cart moves on frictionless wheels, pulled by a rope under tension T. The mass of the cart is m1. A load of mass m2 rests on top of the cart with the coefficient of static friction μs between the cart and the load. The cart is pulled up a ramp that is inclined at angle θ above the horizontal. The rope is parallel to the ramp. What is the maximum tension T that can be applied without causing the load to slip?

Answer 1

`T=(m_1+m_2)g\mu_scos\theta`

Step-by-step explanation:

Using the free body diagram of the load, we obtain according to Newton's laws:

`\sum F_x:f_f-W_x=m_2a(1)\\\sum F_y:N-W_y=0(2)`

In this case we have:

`W_x=m_2gsin\theta\\W_y=m_2gcos\theta`

We have to know the load maximum acceleration in order to calculate the maximum tension. So, we replace
`W_x` and
`W_y` in (1) and (2):

`f_(max)-m_2gsin\theta=m_2a_(max)\\N=m_2gcos\theta\\f_(max)=\mu_s N\\\mu_sm_2gcos\theta-m_2gsin\theta=m_2a_(max)\\a_(max)=\mu_s gcos\theta-gsin\theta`

Now, we use the free body diagram of both bodies. Thus, we have:

`T-W_x=(m_1+m_2)a_(max)\\T-(m_1+m_2)gsin\theta=(m_1+m_2)(\mu_s gcos\theta-gsin\theta)\\T=(m_1+m_2)\mu_s gcos\theta`