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1) gravity of the counterweights will be used to apply a force to the c…

Question

  1. gravity of the counterweights will be used to apply a force to the cart. manipulate and/or erase the arrows to the right to draw the fbd for all the forces acting on the counterweight as the counterweight is falling toward the ground. (assume the wheels on the cart make the friction low enough to ignore.) 2) use the spring scale to measure the magnitude of the weight (fg) for each of the counterweights. this will be the force that will be applied to the cart. (assume friction is low enough to ignore.) 3) manipulate and/or erase the arrows to the right to draw the fbd for all the forces acting on the cart on the table as the counterweight is falling toward the ground. (assume friction is low enough to ignore.)

Explanation:

Step1: Analyze forces on counter - weight

The counter - weight has two forces acting on it as it falls. The force of gravity $F_g$ acts downwards and is given by $F_g = mg$, where $m$ is the mass of the counter - weight and $g$ is the acceleration due to gravity ($g=9.8\ m/s^2$). There is also a tension force $T$ acting upwards (not shown in the initial problem setup but should be in the FBD) which is the force transmitted to the cart. So for the FBD of the counter - weight, draw a downward arrow labeled $F_g$ and an upward arrow labeled $T$.

Step2: Calculate weight of counter - weights

For a $m_1 = 50\ g=0.05\ kg$ counter - weight, using $F_g=mg$, we have $F_{g1}=0.05\times9.8 = 0.49\ N$. For a $m_2 = 100\ g = 0.1\ kg$ counter - weight, $F_{g2}=0.1\times9.8=0.98\ N$.

Step3: Analyze forces on cart

The cart on the table has a normal force $F_N$ acting upwards (equal to its weight if on a horizontal surface), and the tension force $T$ (the same tension as in the counter - weight setup) acting horizontally (the force that makes the cart move). Since friction is ignored, there is no horizontal frictional force. So for the FBD of the cart, draw an upward arrow labeled $F_N$ and a horizontal arrow (in the direction of the pull from the counter - weight) labeled $T$.

Answer:

  1. FBD of counter - weight: Downward arrow $F_g$, upward arrow $T$.
  2. Weight of $50\ g$ counter - weight: $0.49\ N$, weight of $100\ g$ counter - weight: $0.98\ N$.
  3. FBD of cart: Upward arrow $F_N$, horizontal arrow $T$ (in the direction of pull).