Question

1. a 5.88 kg block sits on a ramp that has an angle of inclination of 57.1°. a rope holds the block from moving down the ramp (see diagram). assume there is no friction.

a. draw a free-body diagram of the block.
b. what are the two main action-reaction force pairs acting here?
c. what is the force of tension in the rope?
d. what is the normal force on the block?

Answer:

Part a: Free - body diagram

The block has three forces acting on it:

1. Gravitational force (\(F_g\)): Acts vertically downwards. The magnitude of the gravitational force is given by \(F_g = mg\), where \(m = 5.88\space kg\) and \(g=9.8\space m/s^{2}\).
2. Normal force (\(F_N\)): Acts perpendicular to the surface of the ramp (upwards - perpendicular to the ramp).
3. Tension force (\(F_T\)): Acts along the rope, up the ramp (since it is holding the block from moving down the ramp).

To draw the free - body diagram:

• Represent the block as a point or a square.
• Draw an arrow pointing straight down to represent the gravitational force \(F_g\).
• Draw an arrow perpendicular to the ramp (going out from the block's surface in contact with the ramp) to represent the normal force \(F_N\).
• Draw an arrow along the ramp (upwards) to represent the tension force \(F_T\).

Part b: Action - reaction force pairs

1. Earth - block gravitational force pair: The Earth exerts a gravitational force (pull) on the block (action), and the block exerts an equal - magnitude and opposite - direction gravitational force (pull) on the Earth (reaction). According to Newton's third law, \(F_{Earth\ on\ block}=-F_{block\ on\ Earth}\) (the negative sign indicates opposite direction).
2. Block - ramp normal force pair: The block exerts a normal force (push) on the ramp (action), and the ramp exerts an equal - magnitude and opposite - direction normal force (push) on the block (reaction). So, \(F_{block\ on\ ramp}=-F_{ramp\ on\ block}\)

Part c: Tension force in the rope

Step 1: Analyze the forces along the ramp

Since the block is in equilibrium (not moving), the net force along the ramp is zero. The component of the gravitational force along the ramp is \(F_{g\parallel}=mg\sin\theta\), and the tension force \(F_T\) acts up the ramp. For equilibrium, \(F_T = F_{g\parallel}\)

Step 2: Calculate the component of gravitational force along the ramp

We know that \(m = 5.88\space kg\), \(g = 9.8\space m/s^{2}\) and \(\theta=57.1^{\circ}\)
First, calculate \(\sin(57.1^{\circ})\approx0.84\)
Then, \(F_{g\parallel}=mg\sin\theta=(5.88\space kg)\times(9.8\space m/s^{2})\times\sin(57.1^{\circ})\)
\(F_{g\parallel}=5.88\times9.8\times0.84\)
\(5.88\times9.8 = 57.624\)
\(57.624\times0.84 = 48.40416\space N\approx48.4\space N\)
Since \(F_T=F_{g\parallel}\) (from equilibrium along the ramp), the tension \(F_T\approx48.4\space N\)

Part d: Normal force on the block

Step 1: Analyze the forces perpendicular to the ramp

The block is in equilibrium perpendicular to the ramp, so the net force perpendicular to the ramp is zero. The component of the gravitational force perpendicular to the ramp is \(F_{g\perp}=mg\cos\theta\), and the normal force \(F_N\) acts perpendicular to the ramp (upwards - perpendicular). For equilibrium, \(F_N = F_{g\perp}\)

Step 2: Calculate the component of gravitational force perpendicular to the ramp

We know that \(m = 5.88\space kg\), \(g = 9.8\space m/s^{2}\) and \(\theta = 57.1^{\circ}\)
First, calculate \(\cos(57.1^{\circ})\approx0.55\)
Then, \(F_{g\perp}=mg\cos\theta=(5.88\space kg)\times(9.8\space m/s^{2})\times\cos(57.1^{\circ})\)
\(5.88\times9.8=57.624\)
\(57.624\times0.55 = 31.6932\space N\approx31.7\space N\)
Since \(F_N = F_{g\perp}\) (from equilibrium perpendicular to the ramp), the normal force \(F_N\approx31.7\space N\)

Final Answers

a. (Free - body diagram as described above)
b. 1. Earth pulls block, block pulls Earth (gravitational force pair); 2. Block pushes ramp, ramp pushes block (normal force pair)
c. \(\boldsymbol{\approx48.4\space N}\)
d. \(\boldsymbol{\approx31.7\space N}\)