Question

a string is attached to a block that lies on a horizontal table with a rough surface. in three different trials, the string pulls the block along the table at the same constant velocity to the right. in trial 1, the string is pulled horizontally. in trial 2, the string is pulled at an angle ( \theta_0 ) above the horizontal. in trial 3, the string is pulled at an angle ( \theta_0 ) below the horizontal. in which trial, if any, is the string tension the greatest?
a trial 1
b trial 2
c trial 3
d the string tension is the same in all three trials.

Explanation:

Step1: Analyze forces in each trial

• Trial 1:
• Since the block moves at a constant velocity, the net force is zero.
• Let the tension be \(T_1\), the frictional force \(f_1=\mu N_1\), and \(N_1 = mg\) (normal force equals the weight of the block as there is no vertical component of tension). Also, \(T_1=f_1=\mu mg\) (because \(T_1 - f_1=0\) from horizontal force balance).
• Trial 2:
• Let the tension be \(T_2\).
• Vertical force balance: \(N_2+T_2\sin\theta_0=mg\), so \(N_2 = mg - T_2\sin\theta_0\).
• Horizontal force balance: \(T_2\cos\theta_0=f_2=\mu N_2=\mu(mg - T_2\sin\theta_0)\).
• Rearranging gives \(T_2=\frac{\mu mg}{\cos\theta_0+\mu\sin\theta_0}\).
• Trial 3:
• Let the tension be \(T_3\).
• Vertical force balance: \(N_3=mg + T_3\sin\theta_0\).
• Horizontal force balance: \(T_3\cos\theta_0=f_3=\mu N_3=\mu(mg + T_3\sin\theta_0)\).
• Rearranging gives \(T_3=\frac{\mu mg}{\cos\theta_0-\mu\sin\theta_0}\).

Step2: Compare the tensions

• We have \(T_1=\mu mg\), \(T_2=\frac{\mu mg}{\cos\theta_0+\mu\sin\theta_0}\), and \(T_3=\frac{\mu mg}{\cos\theta_0-\mu\sin\theta_0}\).
• Since \(\cos\theta_0+\mu\sin\theta_0> 1\) (for \(\theta_0>0\) and \(\mu>0\)) and \(\cos\theta_0-\mu\sin\theta_0<1\) (for \(\theta_0>0\) and \(\mu>0\)), we can see that \(T_3>T_1>T_2\).

Answer:

C. Trial 3