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

Answer:

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

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:

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